https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Gamemaster402&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T09:29:06ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_24&diff=1175252020 AMC 10B Problems/Problem 242020-02-09T03:02:46Z<p>Gamemaster402: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
<br />
How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor</math> is the greatest integer not exceeding <math>x</math>.)<br />
<br />
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math><br />
<br />
==Solution==<br />
First notice that the graphs of <math>(x+1000)/70</math> and <math>\sqrt[]{n}</math> intersect at 2 points. Then, notice that <math>(n+1000)/70</math> must be an integer. This means that n is congruent to <math>50 (mod 70)</math>. <br />
<br />
For the first intersection, testing the first few values of <math>n</math> (adding <math>70</math> to <math>n</math> each time and noticing the left side increases by <math>1</math> each time) yields <math>n=20</math> and <math>n=21</math>.<br />
<br />
For the second intersection, using binary search and/or estimating from the graph can narrow down the other cases, being <math>n=47</math>, <math>n=48</math>, <math>n=49</math>, and <math>n=50</math>. This results in a total of 6 cases, for an answer of <math>\boxed{\textbf{(C) }6}</math>.<br />
<br />
~DrJoyo<br />
<br />
==Solution 2 (Graphing)==<br />
<br />
One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of <math>1/70</math>. If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that <math>3</math> values of intersection lay closer to the left side of the stair, and <math>3</math> values lay closer to the right side of the stair.<br />
<br />
With meticulous graphing, you can realize that the answer is <math>\boxed{\textbf{(C) }6}</math>.<br />
<br />
A in-depth graph with intersection points is linked below.<br />
https://www.desmos.com/calculator/e5wk9adbuk<br />
<br />
==Solution 3==<br />
<br />
*Not a reliable or in-depth solution (for the guess and check students)<br />
<br />
We can first consider the equation without a floor function:<br />
<br />
<cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath><br />
<br />
Multiplying both sides by 70 and then squaring:<br />
<br />
<cmath>n^2 + 2000n + 1000000 = 4900n</cmath><br />
<br />
Moving all terms to the left:<br />
<br />
<cmath>n^2 - 2900n + 1000000 = 0</cmath><br />
<br />
Now we can use wishful thinking to determine the factors:<br />
<br />
<cmath>(n-400)(n-2500) = 0</cmath><br />
<br />
This means that for <math>n = 400</math> and <math>n = 2500</math>, the equation will hold without the floor function.<br />
<br />
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:<br />
<br />
For <math>n = 330</math>, left hand side = <math>19</math> but <math>18^2 < 330 < 19^2</math> so right hand side = <math>18</math><br />
<br />
For <math>n = 400</math>, left hand side = <math>20</math> and right hand side = <math>20</math><br />
<br />
For <math>n = 470</math>, left hand side = <math>21</math> and right hand side = <math>21</math><br />
<br />
For <math>n = 540</math>, left hand side = <math>22</math> but <math>540 > 23^2</math> so right hand side = <math>23</math><br />
<br />
Now we move to <math>n = 2500</math><br />
<br />
For <math>n = 2430</math>, left hand side = <math>49</math> and <math>49^2 < 2430 < 50^2</math> so right hand side = <math>49</math><br />
<br />
For <math>n = 2360</math>, left hand side = <math>48</math> and <math>48^2 < 2360 < 49^2</math> so right hand side = <math>48</math><br />
<br />
For <math>n = 2290</math>, left hand side = <math>47</math> and <math>47^2 < 2360 < 48^2</math> so right hand side = <math>47</math><br />
<br />
For <math>n = 2220</math>, left hand side = <math>46</math> but <math>47^2 < 2220</math> so right hand side = <math>47</math><br />
<br />
For <math>n = 2500</math>, left hand side = <math>50</math> and right hand side = <math>50</math><br />
<br />
For <math>n = 2570</math>, left hand side = <math>51</math> but <math>2570 < 51^2</math> so right hand side = <math>50</math><br />
<br />
Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math><br />
<br />
==Solution 4==<br />
<br />
This is my first solution here, so please forgive me for any errors.<br />
<br />
We are given that <cmath>\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor</cmath><br />
<br />
<math>\lfloor\sqrt{n}\rfloor</math> must be an integer, which means that <math>n+1000</math> is divisible by <math>70</math>. As <math>1000\equiv 20\pmod{70}</math>, this means that <math>n\equiv 50\pmod{70}</math>, so we can write <math>n=70k+50</math> for <math>k\in\mathbb{Z}</math>.<br />
<br />
Therefore, <cmath>\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor</cmath><br />
<br />
Also, we can say that <math>\sqrt{70k+50}-1\leq k+15</math> and <math>k+15\leq\sqrt{70k+50}</math><br />
<br />
Squaring the second inequality, we get <math>k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35</math>.<br />
<br />
Similarly solving the first inequality gives us <math>k\leq 19-\sqrt{155}</math> or <math>k\geq 19+\sqrt{155}</math><br />
<br />
<math>\sqrt{155}</math> is slightly larger than <math>12</math>, so instead, we can say <math>k\leq 6</math> or <math>k\geq 32</math>.<br />
<br />
Combining this with <math>5\leq k\leq 35</math>, we get <math>k=5,6,32,33,34,35</math> are all solutions for <math>k</math> that give a valid solution for <math>n</math>, meaning that our answer is <math>\boxed{\textbf{(C)} 6}</math>.<br />
<br />
-Solution By Qqqwerw<br />
<br />
==Solution 5==<br />
<br />
We start with the given equation<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor</cmath>From there, we can start with the general inequality that <math>\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1</math>. This means that<cmath>\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}</cmath>Solving each inequality separately gives us two inequalities:<cmath>n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50</cmath><cmath>n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}</cmath>Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence <math>2+4 = \boxed{\textbf{(C)} 6}</math>.<br />
<br />
~Rekt4<br />
==Video Solution==<br />
On The Spot STEM:<br />
https://youtu.be/BEJybl9TLMA<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2020|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=1165492020 AMC 10A Problems/Problem 132020-02-02T04:19:31Z<p>Gamemaster402: /* Video Solution */</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br />
<br />
==Problem 13==<br />
<br />
A frog sitting at the point <math>(1, 2)</math> begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length <math>1</math>, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices <math>(0,0), (0,4), (4,4),</math> and <math>(4,0)</math>. What is the probability that the sequence of jumps ends on a vertical side of the square<math>?</math><br />
<br />
<math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math><br />
<br />
==Solution==<br />
Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is <math>\frac{1}{4} * 1 = \frac{1}{4}</math>. If the frog goes to the right, it will be in the center of the square at <math>(2,2)</math>, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is <math>\frac{1}{2}</math>. The probability of this happening is <math>\frac{1}{4} * \frac{1}{2} = \frac{1}{8}</math>.<br />
<br />
<br />
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is <math>\frac{1}{2}</math>. Because there's a <math>\frac{1}{2}</math> chance of the frog going up and down, the total probability for this case is <math>\frac{1}{2} * \frac{1}{2} = \frac{1}{4}</math> and summing up all the cases, <math>\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}</math><br />
<br />
==Solution 2==<br />
Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute <math>1 \cdot C + \frac{1}{2} \cdot B</math>. <br />
<br />
<br />
<cmath>1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}</cmath><br />
<cmath>\frac{1}{4} + \frac{3}{8}</cmath><br />
We get <math>\frac{5}{8}</math>, or <math>B</math><br />
<cmath>\text{O O O O O}</cmath> <br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{C A B O O}</cmath><br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{O O O O O}</cmath><br />
-yeskay<br />
<br />
==Solution 3==<br />
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}.}</math><br />
- Lingjun<br />
<br />
==Video Solution==<br />
On The Spot STEM<br />
https://youtu.be/xGs7BjQbGYU<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br />
{{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=1165482020 AMC 10A Problems/Problem 202020-02-02T04:18:50Z<p>Gamemaster402: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
Quadrilateral <math>ABCD</math> satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at point <math>E,</math> and <math>AE=5.</math> What is the area of quadrilateral <math>ABCD?</math><br />
<br />
<math>\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370</math><br />
<br />
== Solution 1 (Just Drop An Altitude)==<br />
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
(I'm very sorry if you're a visual learner)<br />
<br />
~Ultraman<br />
<br />
==Solution 2 (Pro Guessing Strats)==<br />
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus 60, so the answer is <math>\boxed {\textbf{(D) }360}</math>.<br />
<br />
~tigershark22<br />
~(edited by HappyHuman)<br />
<br />
==Solution 3 (coordinates)==<br />
<asy><br />
size(10cm,0);<br />
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br />
draw((-20,0)--(20,0));<br />
draw((0,-15)--(0,35));<br />
draw((10,30)--(-8,-6));<br />
draw(circle((0,0),10));<br />
label("E",(-4.05,-.25),S);<br />
label("D",(10,30),NE);<br />
label("C",(10,0),NE);<br />
label("B",(-8,-6),SW);<br />
label("A",(-10,0),NW);<br />
label("5",(-10,0)--(-5,0), NE);<br />
label("15",(-5,0)--(10,0), N);<br />
label("30",(10,0)--(10,30), E);<br />
dot((-5,0));<br />
dot((-10,0));<br />
dot((-8,-6));<br />
dot((10,0));<br />
dot((10,30));<br />
</asy><br />
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.<br />
<br />
==Solution 4 (Trigonometry)==<br />
Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, and using the fact that <math>cos\angle{AEB}=\frac{1}{\sqrt5}</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>, <math>[ABCD]=[ACD]+\frac{1}{5}[ACD]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
(This solution is incomplete, can someone complete it please-Lingjun)<br />
Latex edited by kc5170<br />
<br />
==Video Solution==<br />
On The Spot STEM<br />
https://www.youtube.com/watch?v=hIdNde2Vln4<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=1165472020 AMC 10A Problems/Problem 202020-02-02T04:18:13Z<p>Gamemaster402: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
Quadrilateral <math>ABCD</math> satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at point <math>E,</math> and <math>AE=5.</math> What is the area of quadrilateral <math>ABCD?</math><br />
<br />
<math>\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370</math><br />
<br />
== Solution 1 (Just Drop An Altitude)==<br />
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
(I'm very sorry if you're a visual learner)<br />
<br />
~Ultraman<br />
<br />
==Solution 2 (Pro Guessing Strats)==<br />
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus 60, so the answer is <math>\boxed {\textbf{(D) }360}</math>.<br />
<br />
~tigershark22<br />
~(edited by HappyHuman)<br />
<br />
==Solution 3 (coordinates)==<br />
<asy><br />
size(10cm,0);<br />
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br />
draw((-20,0)--(20,0));<br />
draw((0,-15)--(0,35));<br />
draw((10,30)--(-8,-6));<br />
draw(circle((0,0),10));<br />
label("E",(-4.05,-.25),S);<br />
label("D",(10,30),NE);<br />
label("C",(10,0),NE);<br />
label("B",(-8,-6),SW);<br />
label("A",(-10,0),NW);<br />
label("5",(-10,0)--(-5,0), NE);<br />
label("15",(-5,0)--(10,0), N);<br />
label("30",(10,0)--(10,30), E);<br />
dot((-5,0));<br />
dot((-10,0));<br />
dot((-8,-6));<br />
dot((10,0));<br />
dot((10,30));<br />
</asy><br />
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.<br />
<br />
==Solution 4 (Trigonometry)==<br />
Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, and using the fact that <math>cos\angle{AEB}=\frac{1}{\sqrt5}</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>, <math>[ABCD]=[ACD]+\frac{1}{5}[ACD]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
(This solution is incomplete, can someone complete it please-Lingjun)<br />
Latex edited by kc5170<br />
<br />
==Video Solution==<br />
On The Spot STEM<br />
https://www.youtube.com/watch?v=xGs7BjQbGYU<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_24&diff=1165462020 AMC 12A Problems/Problem 242020-02-02T04:17:04Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>==Problem 24==<br />
Suppose that <math>\triangle{ABC}</math> is an equilateral triangle of side length <math>s</math>, with the property that there is a unique point <math>P</math> inside the triangle such that <math>AP=1</math>, <math>BP=\sqrt{3}</math>, and <math>CP=2</math>. What is <math>s</math>?<br />
<br />
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}</math><br />
<br />
==Solution==<br />
<br />
<asy><br />
draw((0,0)--(4,5.65)--(8,0)--cycle);<br />
label("A", (4,5.65), N, p = fontsize(10pt));<br />
label("C", (8,0), SE, p = fontsize(10pt));<br />
label("B", (0,0), SW, p = fontsize(10pt));<br />
label("P", (3.5,3.5), NW, p = fontsize(10pt));<br />
draw((0,0)--(3.5,3.5));<br />
label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE);<br />
draw((8,0)--(3.5,3.5));<br />
label("2",(8,0)--(3.5,3.5), SW);<br />
draw((4,5.65)--(3.5,3.5));<br />
label("1",(4,5.65)--(3.5,3.5), E);<br />
draw((8,0)--(4,5.65)--(11,5.65)--cycle);<br />
label("$P'$", (6,4.75), NE, p = fontsize(10pt));<br />
draw((4,5.65)--(6,4.75));<br />
label("1",(4,5.65)--(6,4.75), S);<br />
draw((8,0)--(6,4.75));<br />
label("$\sqrt{3}$",(8,0)--(6,4.75), E);<br />
draw((3.5,3.5)--(6,4.75));<br />
label("1", (3.5,3.5)--(6,4.75), SE);<br />
<br />
</asy><br />
<br />
We begin by rotating <math>\triangle{ABC}</math> by <math>60^{\circ}</math> about <math>A</math>, such that in <math>\triangle{A'B'C'}</math>, <math>B' = C</math>. We see that<br />
<math>\triangle{APP'}</math> is equilateral with side length <math>1</math>, meaning that <math>\angle APP' = 60^{\circ}</math>. We also see that <math>\triangle{CPP'}</math> is a <math>30-60-90</math> right triangle, meaning that <math>\angle CPP'= 60^{\circ}</math>. Thus, by adding the two together, we see that <math>\angle APC = 120^{\circ}</math>. We can now use the law of cosines as following:<br />
<br />
<cmath>s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}}</cmath><br />
<cmath>s^2 = 1 + 4 - 2(1)(2)\cos{120^{\circ}}</cmath><br />
<cmath>s^2 = 5 - 4(-\frac{1}{2})</cmath><br />
<cmath>s = \sqrt{5 + 2}</cmath><br />
<br />
giving us that <math>s = \boxed{\textbf{(B) } \sqrt{7}}</math>. ~ciceronii<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=mUW4zcrRL54<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_15&diff=1099732017 AMC 10A Problems/Problem 152019-09-26T15:05:42Z<p>Gamemaster402: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
Chloé chooses a real number uniformly at random from the interval <math>[0, 2017]</math>. Independently, Laurent chooses a real number uniformly at random from the interval <math>[0, 4034]</math>. What is the probability that Laurent's number is greater than Chloé's number? (Assume they cannot be equal)<br />
<br />
<math> \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}</math><br />
<br />
==Solution 1==<br />
Denote "winning" to mean "picking a greater number".<br />
There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>[2017, 4034]</math>. In this case, Chloé cannot possibly win, since the maximum number she can pick is <math>2017</math>. Otherwise, if Laurent picks a number in the interval <math>[0, 2017]</math>, with probability <math>\frac{1}{2}</math>, then the two people are symmetric, and each has a <math>\frac{1}{2}</math> chance of winning. Then, the total probability is <math>\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}</math><br />
<br />
==Solution 2==<br />
We can use geometric probability to solve this.<br />
Suppose a point <math>(x,y)</math> lies in the <math>xy</math>-plane. Let <math>x</math> be Chloe's number and <math>y</math> be Laurent's number. Then obviously we want <math>y>x</math>, which basically gives us a region above a line. We know that Chloe's number is in the interval <math>[0,2017]</math> and Laurent's number is in the interval <math>[0,4034]</math>, so we can create a rectangle in the plane, whose length is <math>2017</math> and whose width is <math>4034</math>. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from <math>1</math>. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line <math>y>x</math>, which is <math>\frac{2017 \cdot 2017}{2}</math>. Instead of bashing this out we know that the rectangle has area <math>2017 \cdot 4034</math>. So the probability that Laurent has a smaller number is <math>\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}</math>. Simplifying the expression yields <math>\frac{1}{4}</math> and so <math>1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}</math>.<br />
<br />
==Solution 3==<br />
Scale down by <math>2017</math> to get that Chloe picks from <math>[0,1]</math> and Laurent picks from <math>[0,2]</math>. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of <math>0.5</math>. Therefore, Laurent has a range of <math>0.5</math> to <math>2</math> to pick from, on average, which is a length of <math>2-0.5=1.5</math> out of a total length of <math>2-0=2</math>. Therefore, the probability is <math>1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}</math><br />
==Video Solution==<br />
A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Probability Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_15&diff=1086702010 AIME I Problems/Problem 152019-08-10T06:45:50Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>__TOC__<br />
== Problem ==<br />
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Let <math>p</math> and <math>q</math> be positive [[relatively prime]] integers such that <math>\frac {AM}{CM} = \frac {p}{q}</math>. Find <math>p + q</math>.<br />
<br />
== Solution ==<br />
<center><asy> /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);<br />
<br />
/* segments and figures */<br />
draw((0,0)--(15,0));<br />
draw((15,0)--(6.66667,9.97775));<br />
draw((6.66667,9.97775)--(0,0));<br />
draw((7.33333,0)--(6.66667,9.97775));<br />
draw(circle((4.66667,2.49444),2.49444));<br />
draw(circle((9.66667,2.49444),2.49444));<br />
draw((4.66667,0)--(4.66667,2.49444));<br />
draw((9.66667,2.49444)--(9.66667,0));<br />
<br />
/* points and labels */<br />
label("r",(10.19662,1.92704),SE);<br />
label("r",(5.02391,1.8773),SE);<br />
dot((0,0));<br />
label("$A$",(-1.04408,-0.60958),NE);<br />
dot((15,0));<br />
label("$C$",(15.41907,-0.46037),NE);<br />
dot((6.66667,9.97775));<br />
label("$B$",(6.66525,10.23322),NE);<br />
label("$15$",(6.01866,-1.15669),NE);<br />
label("$13$",(11.44006,5.50815),NE);<br />
label("$12$",(2.28834,5.75684),NE);<br />
dot((7.33333,0));<br />
label("$M$",(7.56053,-0.908),NE);<br />
dot((4.66667,2.49444));<br />
label("$I_1$",(3.97942,2.92179),NE);<br />
dot((9.66667,2.49444));<br />
label("$I_2$",(10.04741,2.97153),NE);<br />
clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
Let <math>AM = x</math>, then <math>CM = 15 - x</math>. Also let <math>BM = d</math> Clearly, <math>\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}</math>. We can also express each area by the rs formula. Then <math>\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}</math>. Equating and cross-multiplying yields <math>25x + 2dx = 15d + 180</math> or <math>d = \frac {25x - 180}{15 - 2x}.</math> Note that for <math>d</math> to be positive, we must have <math>7.2 < x < 7.5</math>.<br />
<br />
By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math><br />
<br />
''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math>12x</math> is an integer. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''<br />
<br />
Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>.<br />
<br />
=== Solution 2 ===<br />
Let <math>AM = 2x</math> and <math>BM = 2y</math> so <math>CM = 15 - 2x</math>. Let the [[incenter]]s of <math>\triangle ABM</math> and <math>\triangle BCM</math> be <math>I_1</math> and <math>I_2</math> respectively, and their equal inradii be <math>r</math>. From <math>r = \sqrt {(s - a)(s - b)(s - c)/s}</math>, we find that<br />
<br />
<cmath>\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\<br />
& = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}</cmath><br />
<br />
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields<br />
<br />
<cmath>\begin{align*}<br />
2y^2 - 30 = 2xy + 5x - 7y \\<br />
2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}<br />
</cmath><br />
<br />
and adding these we have<br />
<br />
<cmath><br />
4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\<br />
\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.<br />
</cmath><br />
<br />
=== Solution 3 ===<br />
Let the incircle of <math>ABM</math> hit <math>AM</math>, <math>AB</math>, <math>BM</math> at <math>X_{1},Y_{1},Z_{1}</math>, and let the incircle of <math>CBM</math> hit <math>MC</math>, <math>BC</math>, <math>BM</math> at <math>X_{2},Y_{2},Z_{2}</math>. Draw the incircle of <math>ABC</math>, and let it be tangent to <math>AC</math> at <math>X</math>. Observe that we have a homothety centered at A sending the incircle of <math>ABM</math> to that of <math>ABC</math>, and one centered at <math>C</math> taking the incircle of <math>BCM</math> to that of <math>ABC</math>. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is <math>AX_{1}/AX=CX_{2}/CX</math>.<br />
<br />
By standard computations, <math>AX=\dfrac{AB+AC-BC}{2}=7</math> and <math>CX=\dfrac{BC+AC-AB}{2}=8</math>. Now, let <math>AX_{1}=7x</math> and <math>CX_{2}=8x</math>. We will now go around and chase lengths. Observe that <math>BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x</math>. Then, <math>BZ_{1}=12-7x</math>. We also have <math>CY_{2}=CX_{2}=8x</math>, so <math>BY_{2}=13-8x</math> and <math>BZ_{2}=13-8x</math>.<br />
<br />
Observe now that <math>X_{1}M+MX_{2}=AC-15x=15(1-x)</math>. Also,<math>X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)</math>. Solving, we get <math>X_{1}M=8-8x</math> and <math>MX_{2}=7-7x</math> (as a side note, note that <math>AX_{1}+MX_{2}=X_{1}M+X_{2}C</math>, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).<br />
<br />
Now, we get <math>BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x</math>. To finish, we will compute area ratios. <math>\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}</math>. Also, since their inradii are equal, we get <math>\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}</math>. Equating and cross multiplying yields the quadratic <math>3x^{2}-8x+4=0</math>, so <math>x=2/3,2</math>. However, observe that <math>AX_{1}+CX_{2}=15x<15</math>, so we take <math>x=2/3</math>. Our ratio is therefore <math>\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}</math>, giving the answer <math>\boxed{045}</math>.<br />
<br />
Note: Once we have <math>MX_1=8-8x</math> and <math>MX_2=7-7x</math>, it's bit easier to use use the right triangle of <math>O_1MO_2</math> than chasing the area ratio. The inradius of <math>\triangle{ABC}</math> can be calculated to be <math>r=\sqrt{14}</math>, and the inradius of <math>ABM</math> and <math>ACM</math> are <math>r_1=r_2= xr</math>, so, <br />
<cmath> O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2</cmath><br />
or,<br />
<cmath> (15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2 </cmath><br />
<cmath>112(1-x)^2 = 28x^2</cmath><br />
<cmath>4(1-x)^2 = x^2</cmath><br />
We get <math>x=\frac{2}{3}</math> or <math>x=2</math>.<br />
<br />
=== Solution 4 ===<br />
Suppose the incircle of <math>ABM</math> touches <math>AM</math> at <math>X</math>, and the incircle of <math>CBM</math> touches <math>CM</math> at <math>Y</math>. Then<br />
<br />
<cmath>r = AX \tan(A/2) = CY \tan(C/2)</cmath><br />
<br />
We have <math>\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}</math>, <math>\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}</math><br />
<br />
<math>\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}</math>, <math>\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}</math>,<br />
<br />
Therefore <math>AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.</math><br />
<br />
And since <math>AX=\frac{1}{2}(12+AM-BM)</math>, <math>CY = \frac{1}{2}(13+CM-BM)</math>, <br />
<br />
<cmath> \frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}</cmath><br />
<br />
<cmath> 96+8AM-8BM = 91 +7CM-7BM</cmath><br />
<br />
<cmath>BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)</cmath><br />
<br />
Now,<br />
<br />
<math>\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}</math><br />
<br />
<cmath>\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}</cmath><br />
<cmath>6AM^2 - 35AM = 45AM-264</cmath><br />
<cmath>3AM^2 -40AM+132=0</cmath><br />
<cmath>(3AM-22)(AM-6)=0</cmath><br />
<br />
So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>.<br />
<br />
'''Sidenote'''<br />
<br />
In the problem, <math>BM=10</math> and the equal inradius of the two triangles happens to be <math> \frac {2\sqrt{14}}{3}</math>.<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0<br />
<br />
== See Also ==<br />
<br />
<br />
{{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_13&diff=1081622019 AIME II Problems/Problem 132019-08-02T20:32:37Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>==Problem==<br />
Regular octagon <math>A_1A_2A_3A_4A_5A_6A_7A_8</math> is inscribed in a circle of area <math>1.</math> Point <math>P</math> lies inside the circle so that the region bounded by <math>\overline{PA_1},\overline{PA_2},</math> and the minor arc <math>\widehat{A_1A_2}</math> of the circle has area <math>\tfrac{1}{7},</math> while the region bounded by <math>\overline{PA_3},\overline{PA_4},</math> and the minor arc <math>\widehat{A_3A_4}</math> of the circle has area <math>\tfrac{1}{9}.</math> There is a positive integer <math>n</math> such that the area of the region bounded by <math>\overline{PA_6},\overline{PA_7},</math> and the minor arc <math>\widehat{A_6A_7}</math> of the circle is equal to <math>\tfrac{1}{8}-\tfrac{\sqrt2}{n}.</math> Find <math>n.</math><br />
<br />
==Solution==<br />
This problem is not difficult, but the calculation is tormenting.<br />
<br />
The actual size of the diagram doesn't matter. <br />
To make calculation easier, we discard the original area of the circle, <math>1</math>, and assume the side length of the octagon is <math>2</math><br />
<br />
Let <math>r</math> denotes the radius of the circle, <math>O</math> be the center of the circle.<br />
<br />
<math>r^2= 1^2 + (\sqrt{2}+1)^2= 4+2\sqrt{2}</math><br />
<br />
Now, we need to find the "D"shape, the small area enclosed by one side of the octagon and 1/8 of the circumference of the circle<br />
<br />
<math>D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1)</math><br />
<br />
Let<br />
<math>PU</math> be the height of <math>\triangle A_1 A_2 P</math>,<br />
<math>PV</math> be the height of <math>\triangle A_3 A_4 P</math>,<br />
<math>PW</math> be the height of <math>\triangle A_6 A_7 P</math>,<br />
<br />
From the 1/7 and 1/9 condition<br />
<br />
we have<br />
<br />
<math> \triangle P A_1 A_2= \frac{\pi r^2}{7} - D= \frac{1}{7} \pi (4+2\sqrt{2})-(\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1))</math><br />
<br />
<math> \triangle P A_3 A_4= \frac{\pi r^2}{9} - D= \frac{1}{9} \pi (4+2\sqrt{2})-(\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1))</math><br />
<br />
which gives<br />
<br />
<math>PU= (\frac{1}{7}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1</math><br />
<br />
<math>PV= (\frac{1}{9}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1</math><br />
<br />
<br />
Now, let <math>A_1 A_2</math> intersects <math>A_3 A_4</math> at <math>X</math>, <math>A_1 A_2</math> intersects <math>A_6 A_7</math> at <math>Y</math>,<math>A_6 A_7</math> intersects <math>A_3 A_4</math> at <math>Z</math><br />
<br />
Clearly, <math>\triangle XYZ</math> is an isosceles right triangle, with right angle at <math>X</math><br />
<br />
and the height with regard to which shall be <math>3+2\sqrt2</math><br />
<br />
That <math>\frac{PU}{\sqrt{2}} + \frac{PV}{\sqrt{2}} + PW = 3+2\sqrt2</math> is a common sense<br />
<br />
which gives <math>PW= 3+2\sqrt2-\frac{PU}{\sqrt{2}} - \frac{PV}{\sqrt{2}}</math><br />
<br />
<math>=3+2\sqrt{2}-\frac{1}{\sqrt{2}}((\frac{1}{7}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1+(\frac{1}{9}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1))</math><br />
<br />
<math>=1+\sqrt{2}- \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi(4+2\sqrt{2})</math><br />
<br />
Now, we have the area for <math>D</math> and the area for <math>\triangle P A_6 A_7</math><br />
<br />
we add them together<br />
<br />
<math>Target Area= \frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1) + (1+\sqrt{2})- \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi(4+2\sqrt{2})</math><br />
<br />
<math>=(\frac{1}{8} - \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}))Total Area</math><br />
<br />
The answer should therefore be <math>\frac{1}{8}- \frac{\sqrt{2}}{2}(\frac{16}{63}-\frac{16}{64})=\frac{1}{8}- \frac{\sqrt{2}}{504}</math><br />
<br />
The final answer is, therefore, <math>\boxed{504}</math><br />
<br />
-By SpecialBeing2017<br />
<br />
==Video Solution==<br />
https://youtu.be/B_Drjjn0vv0<br />
<br />
==Solution 2==<br />
<br />
Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point <math>P</math> from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram. Note that the area bounded by <math>\overline{A_i}{A_j}</math> and the arc <math>A_iA_j</math> is fixed, so we only need to consider the relevant triangles.<br />
<br />
<asy><br />
size(7cm);<br />
draw(Circle((0,0),1));<br />
<br />
pair P = (0.1,-0.15);<br />
filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red);<br />
filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red);<br />
filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);<br />
<br />
dot(P);<br />
<br />
for(int i=0; i<8; ++i)<br />
{<br />
draw(dir(22.5+45i)--dir(67.5+45i));<br />
draw((0,0)--dir(22.5+45i),gray+dashed);<br />
}<br />
<br />
draw(dir(135)--dir(-45),blue+linewidth(1));<br />
<br />
label("$P$", P, dir(-75));<br />
<br />
<br />
<br />
label("$A_1$", dir(112.5), dir(112.5));<br />
label("$A_2$", dir(112.5-45), dir(112.5-45));<br />
label("$A_3$", dir(112.5-90), dir(112.5-90));<br />
label("$A_4$", dir(112.5-135), dir(112.5-135));<br />
label("$A_5$", dir(112.5-180), dir(112.5-180));<br />
label("$A_6$", dir(112.5-225), dir(112.5-225));<br />
label("$A_7$", dir(112.5-270), dir(112.5-270));<br />
label("$A_8$", dir(112.5-315), dir(112.5-315));<br />
<br />
dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315));<br />
</asy><br />
<br />
Define one arbitrary unit as the distance that you need to move <math>P</math> from <math>A_1A_2</math> to change the area of <math>\triangle PA_1A_2</math> by <math>1</math>. We can see that <math>P</math> was moved down by <math>\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}</math> units to make the area defined by <math>P</math>, <math>A_1</math>, and <math>A_2</math> <math>\tfrac{1}{7}</math>. Similarly, <math>P</math> was moved right by <math>\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}</math> to make the area defined by <math>P</math>, <math>A_3</math>, and <math>A_4</math> <math>\tfrac{1}{9}</math>. This means that <math>P</math> has coordinates <math>(\tfrac{1}{72},-\tfrac{1}{56})</math>.<br />
<br />
Now, we need to consider how this displacement in <math>P</math> affected the area defined by <math>P</math>, <math>A_6</math>, and <math>A_7</math>. This is equivalent to finding the shortest distance between <math>P</math> and the blue line in the diagram (as <math>K=\tfrac{1}{2}bh</math> and the blue line represents <math>h</math> while <math>b</math> is fixed). Using an isosceles right triangle, one can find the that shortest distance between <math>P</math> and this line is <math>\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}</math>.<br />
<br />
Remembering the definition of our unit, this yields a final area of<br />
<cmath>\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}}.</cmath><br />
<br />
-Archeon<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=1081152019 AMC 10B Problems/Problem 222019-08-01T06:50:34Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br />
<br />
==Problem==<br />
<br />
Raashan, Sylvia, and Ted play the following game. Each starts with <math> \$1</math>. A bell rings every <math>15</math> seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives <math>\$1</math> to that player. What is the probability that after the bell has rung <math>2019</math> times, each player will have <math>\$1</math>? (For example, Raashan and Ted may each decide to give <math>\$1</math> to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have <math>\$0</math>, Sylvia will have <math>\$2</math>, and Ted will have <math>\$1</math>, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their <math> \$1</math> to, and the holdings will be the same at the end of the second round.)<br />
<br />
<math>\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}</math><br />
<br />
==Solution==<br />
On the first turn, each player starts off with <math>\$1</math>. Each turn after that, there are only two possibilities: either everyone stays at <math>\$1</math>, which we will write as <math>(1-1-1)</math>, or the distribution of money becomes <math>\$2-\$1-\$0</math> in some order, which we write as <math>(2-1-0)</math>. We will consider these two states separately.<br />
<br />
In the <math>(1-1-1)</math> state, each person has two choices for whom to give their dollar to, meaning there are <math>2^3=8</math> possible ways that the money can be rearranged. Note that there are only two ways that we can reach <math>(1-1-1)</math> again: <br />
<br />
1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.<br />
<br />
2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.<br />
<br />
Thus, the probability of staying in the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of going to the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math> (we can check that the 6 other possibilities lead to <math>(2-1-0)</math>)<br />
<br />
<br />
In the <math>(2-1-0)</math> state, we will label the person with <math>\$2</math> as person A, the person with <math>\$1</math> as person B, and the person with <math>\$0</math> as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total <math>2\cdot 2 = 4</math> ways the money can be redistributed. The only way that the distribution can return to <math>(1-1-1)</math> is if A gives <math>\$1</math> to B, and B gives <math>\$1</math> to C. We check the other possibilities to find that they all lead back to <math>(2-1-0)</math>. Thus, the probability of going to the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of staying in the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math>.<br />
<br />
No matter which state we are in, the probability of going to the <math>(1-1-1)</math> state is always <math>\frac{1}{4}</math>. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a <math>\frac{1}{4}</math> probability of going to the <math>(1-1-1)</math> state after the 2019th bell ring. Thus, our answer is simply <math>\boxed{\textbf{(B) } \frac{1}{4}}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/XT440PjAFmQ<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br />
{{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=1079272019 AMC 10B Problems/Problem 212019-07-25T21:06:55Z<p>Gamemaster402: /* Solution 2 (Easier) */</p>
<hr />
<div>==Problem==<br />
<br />
Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br />
<br />
<math>\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}</math><br />
<br />
==Solution==<br />
<br />
We firstly want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the seecond head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is <math>THTHH</math>, which has a probability of <math>\frac{1}{2^5} = \frac{1}{32}</math>. Furthermore, she can prolong her coin flipping by adding an extra <math>TH</math>, which itself has a probability of <math>\frac{1}{2^2} = \frac{1}{4}</math>. Since she can do this indefinitely, this gives an infinite geometric series, which means the answer (by the geometric series sum formula) is <math>\frac{\frac{1}{32}}{1-\frac{1}{4}} = \boxed{\textbf{(B) }\frac{1}{24}}</math>.<br />
<br />
==Solution 2 (Easier)==<br />
Note that the sequence must start in THT, which happens with <math>\frac{1}{8}</math> probability. Now, let <math>P</math> be the probability that Debra will get two heads in a row after flipping THT. Either Debra flips two heads in a row immediately (probability <math>\frac{1}{4}</math>), or flips a head and then a tail and reverts back to the "original position" (probability <math>\frac{1}{4}P</math>). Therefore, <math>P=\frac{1}{4}+\frac{1}{4}P</math>, so <math>P=\frac{1}{3}</math>, so our final answer is <math>\frac{1}{8}\times\frac{1}{3}=\boxed{\textbf{(B) }\frac{1}{24}}</math>. -Stormersyle get rect<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=2f1zEvfUe9o<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_8&diff=1075112018 AIME II Problems/Problem 82019-07-09T01:46:07Z<p>Gamemaster402: /* Video Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A frog is positioned at the origin of the coordinate plane. From the point <math>(x, y)</math>, the frog can jump to any of the points <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, or <math>(x, y + 2)</math>. Find the number of distinct sequences of jumps in which the frog begins at <math>(0, 0)</math> and ends at <math>(4, 4)</math>.<br />
<br />
==Solution 1==<br />
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to <math>(4,4)</math> in one move are <math>(2,4),(3,4),(4,2),</math> and <math>(4,3)</math>. This applies to any other point, thus we can work our way from <math>(0,0)</math> to <math>(4,4)</math>, recording down the number of ways to get to each point recursively. <br />
<br />
<math>(0,0): 1</math><br />
<br />
<math>(1,0)=(0,1)=1</math><br />
<br />
<math>(2,0)=(0, 2)=2</math><br />
<br />
<math>(3,0)=(0, 3)=3</math><br />
<br />
<math>(4,0)=(0, 4)=5</math><br />
<br />
<math>(1,1)=2</math>, <math>(1,2)=(2,1)=5</math>, <math>(1,3)=(3,1)=10</math>, <math>(1,4)=(4,1)= 20</math><br />
<br />
<math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71</math><br />
<br />
<math>(3,3)=84, (3,4)=(4,3)=207</math><br />
<br />
<math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}</math><br />
<br />
A diagram of the numbers:<br />
<br />
5 - 20 - 71 - 207 - <math>\boxed{556}</math><br />
<br />
3 - 10 - 32 - 84 - 207<br />
<br />
2 - 5 - 14 - 32 - 71<br />
<br />
1 - 2 - 5 - 10 - 20<br />
<br />
1 - 1 - 2 - 3 - 5<br />
<br />
==Solution 2==<br />
We'll refer to the moves <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, and <math>(x, y + 2)</math> as <math>R_1</math>, <math>R_2</math>, <math>U_1</math>, and <math>U_2</math>, respectively. Then the possible sequences of moves that will take the frog from <math>(0,0)</math> to <math>(4,4)</math> are all the permutations of <math>U_1U_1U_1U_1R_1R_1R_1R_1</math>, <math>U_2U_1U_1R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_1R_1</math>, <math>U_2U_2R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_2</math>, <math>U_2U_2R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_2</math>, and <math>U_2U_2R_2R_2</math>. We can reduce the number of cases using symmetry.<br />
<br />
Case 1: <math>U_1U_1U_1U_1R_1R_1R_1R_1</math><br />
<br />
There are <math>\frac{8!}{4!4!} = 70</math> possibilities for this case.<br />
<br />
Case 2: <math>U_2U_1U_1R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>2 \cdot \frac{7!}{4!2!} = 210</math> possibilities for this case.<br />
<br />
Case 3: <math>U_2U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>\frac{6!}{2!2!} = 180</math> possibilities for this case.<br />
<br />
Case 4: <math>U_2U_2R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{6!}{2!4!} = 30</math> possibilities for this case.<br />
<br />
Case 5: <math>U_2U_2R_2R_1R_1</math> or <math>U_2U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{5!}{2!2!} = 60</math> possibilities for this case.<br />
<br />
Case 6: <math>U_2U_2R_2R_2</math><br />
<br />
There are <math>\frac{4!}{2!2!} = 6</math> possibilities for this case.<br />
<br />
Adding up all these cases gives us <math>70+210+180+30+60+6=\boxed{556}</math> ways.<br />
<br />
==Solution 3 (General Case)==<br />
<br />
Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these <math>4</math> points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath><br />
<br />
Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath><br />
<br />
<cmath>\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)</cmath><br />
<br />
<cmath>\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)</cmath><br />
<br />
<cmath>\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)</cmath><br />
<br />
<cmath>\varphi (1,1)=\varphi (1,0)+\varphi (0,1)</cmath><br />
<br />
Start with <math>\varphi (0,0)=1</math>, use this method and draw the figure below, we can finally get <cmath>\varphi (4,4)=556</cmath> (In order to make the LaTeX thing more beautiful to look at, I put <math>0</math> to make every number a <math>3</math>-digits number)<br />
<br />
<cmath>005-020-071-207-\boxed{556}</cmath> <cmath>003-010-032-084-207</cmath> <cmath>002-005-014-032-071</cmath> <cmath>001-002-005-010-020</cmath> <cmath>001-001-002-003-005</cmath><br />
<br />
So the total number of distinct sequences of jumps for the frog to reach <math>(4,4)</math> is <math>\boxed {556}</math>.<br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Video Solution==<br />
<br />
On The Spot STEM : <br />
https://www.youtube.com/watch?v=v2fo3CaAhmM<br />
<br />
{{AIME box|year=2018|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_8&diff=1075102018 AIME II Problems/Problem 82019-07-09T01:45:58Z<p>Gamemaster402: /* Video Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A frog is positioned at the origin of the coordinate plane. From the point <math>(x, y)</math>, the frog can jump to any of the points <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, or <math>(x, y + 2)</math>. Find the number of distinct sequences of jumps in which the frog begins at <math>(0, 0)</math> and ends at <math>(4, 4)</math>.<br />
<br />
==Solution 1==<br />
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to <math>(4,4)</math> in one move are <math>(2,4),(3,4),(4,2),</math> and <math>(4,3)</math>. This applies to any other point, thus we can work our way from <math>(0,0)</math> to <math>(4,4)</math>, recording down the number of ways to get to each point recursively. <br />
<br />
<math>(0,0): 1</math><br />
<br />
<math>(1,0)=(0,1)=1</math><br />
<br />
<math>(2,0)=(0, 2)=2</math><br />
<br />
<math>(3,0)=(0, 3)=3</math><br />
<br />
<math>(4,0)=(0, 4)=5</math><br />
<br />
<math>(1,1)=2</math>, <math>(1,2)=(2,1)=5</math>, <math>(1,3)=(3,1)=10</math>, <math>(1,4)=(4,1)= 20</math><br />
<br />
<math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71</math><br />
<br />
<math>(3,3)=84, (3,4)=(4,3)=207</math><br />
<br />
<math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}</math><br />
<br />
A diagram of the numbers:<br />
<br />
5 - 20 - 71 - 207 - <math>\boxed{556}</math><br />
<br />
3 - 10 - 32 - 84 - 207<br />
<br />
2 - 5 - 14 - 32 - 71<br />
<br />
1 - 2 - 5 - 10 - 20<br />
<br />
1 - 1 - 2 - 3 - 5<br />
<br />
==Solution 2==<br />
We'll refer to the moves <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, and <math>(x, y + 2)</math> as <math>R_1</math>, <math>R_2</math>, <math>U_1</math>, and <math>U_2</math>, respectively. Then the possible sequences of moves that will take the frog from <math>(0,0)</math> to <math>(4,4)</math> are all the permutations of <math>U_1U_1U_1U_1R_1R_1R_1R_1</math>, <math>U_2U_1U_1R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_1R_1</math>, <math>U_2U_2R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_2</math>, <math>U_2U_2R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_2</math>, and <math>U_2U_2R_2R_2</math>. We can reduce the number of cases using symmetry.<br />
<br />
Case 1: <math>U_1U_1U_1U_1R_1R_1R_1R_1</math><br />
<br />
There are <math>\frac{8!}{4!4!} = 70</math> possibilities for this case.<br />
<br />
Case 2: <math>U_2U_1U_1R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>2 \cdot \frac{7!}{4!2!} = 210</math> possibilities for this case.<br />
<br />
Case 3: <math>U_2U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>\frac{6!}{2!2!} = 180</math> possibilities for this case.<br />
<br />
Case 4: <math>U_2U_2R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{6!}{2!4!} = 30</math> possibilities for this case.<br />
<br />
Case 5: <math>U_2U_2R_2R_1R_1</math> or <math>U_2U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{5!}{2!2!} = 60</math> possibilities for this case.<br />
<br />
Case 6: <math>U_2U_2R_2R_2</math><br />
<br />
There are <math>\frac{4!}{2!2!} = 6</math> possibilities for this case.<br />
<br />
Adding up all these cases gives us <math>70+210+180+30+60+6=\boxed{556}</math> ways.<br />
<br />
==Solution 3 (General Case)==<br />
<br />
Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these <math>4</math> points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath><br />
<br />
Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath><br />
<br />
<cmath>\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)</cmath><br />
<br />
<cmath>\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)</cmath><br />
<br />
<cmath>\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)</cmath><br />
<br />
<cmath>\varphi (1,1)=\varphi (1,0)+\varphi (0,1)</cmath><br />
<br />
Start with <math>\varphi (0,0)=1</math>, use this method and draw the figure below, we can finally get <cmath>\varphi (4,4)=556</cmath> (In order to make the LaTeX thing more beautiful to look at, I put <math>0</math> to make every number a <math>3</math>-digits number)<br />
<br />
<cmath>005-020-071-207-\boxed{556}</cmath> <cmath>003-010-032-084-207</cmath> <cmath>002-005-014-032-071</cmath> <cmath>001-002-005-010-020</cmath> <cmath>001-001-002-003-005</cmath><br />
<br />
So the total number of distinct sequences of jumps for the frog to reach <math>(4,4)</math> is <math>\boxed {556}</math>.<br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Video Solution==<br />
<br />
On The Spot STEM : <br />
https://www.youtube.com/watch?v=v2fo3CaAhmM&t=11s<br />
<br />
{{AIME box|year=2018|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_8&diff=1075092018 AIME II Problems/Problem 82019-07-09T01:45:51Z<p>Gamemaster402: /* Solution 3 (General Case) */</p>
<hr />
<div>==Problem==<br />
<br />
A frog is positioned at the origin of the coordinate plane. From the point <math>(x, y)</math>, the frog can jump to any of the points <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, or <math>(x, y + 2)</math>. Find the number of distinct sequences of jumps in which the frog begins at <math>(0, 0)</math> and ends at <math>(4, 4)</math>.<br />
<br />
==Solution 1==<br />
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to <math>(4,4)</math> in one move are <math>(2,4),(3,4),(4,2),</math> and <math>(4,3)</math>. This applies to any other point, thus we can work our way from <math>(0,0)</math> to <math>(4,4)</math>, recording down the number of ways to get to each point recursively. <br />
<br />
<math>(0,0): 1</math><br />
<br />
<math>(1,0)=(0,1)=1</math><br />
<br />
<math>(2,0)=(0, 2)=2</math><br />
<br />
<math>(3,0)=(0, 3)=3</math><br />
<br />
<math>(4,0)=(0, 4)=5</math><br />
<br />
<math>(1,1)=2</math>, <math>(1,2)=(2,1)=5</math>, <math>(1,3)=(3,1)=10</math>, <math>(1,4)=(4,1)= 20</math><br />
<br />
<math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71</math><br />
<br />
<math>(3,3)=84, (3,4)=(4,3)=207</math><br />
<br />
<math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}</math><br />
<br />
A diagram of the numbers:<br />
<br />
5 - 20 - 71 - 207 - <math>\boxed{556}</math><br />
<br />
3 - 10 - 32 - 84 - 207<br />
<br />
2 - 5 - 14 - 32 - 71<br />
<br />
1 - 2 - 5 - 10 - 20<br />
<br />
1 - 1 - 2 - 3 - 5<br />
<br />
==Solution 2==<br />
We'll refer to the moves <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, and <math>(x, y + 2)</math> as <math>R_1</math>, <math>R_2</math>, <math>U_1</math>, and <math>U_2</math>, respectively. Then the possible sequences of moves that will take the frog from <math>(0,0)</math> to <math>(4,4)</math> are all the permutations of <math>U_1U_1U_1U_1R_1R_1R_1R_1</math>, <math>U_2U_1U_1R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_1R_1</math>, <math>U_2U_2R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_2</math>, <math>U_2U_2R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_2</math>, and <math>U_2U_2R_2R_2</math>. We can reduce the number of cases using symmetry.<br />
<br />
Case 1: <math>U_1U_1U_1U_1R_1R_1R_1R_1</math><br />
<br />
There are <math>\frac{8!}{4!4!} = 70</math> possibilities for this case.<br />
<br />
Case 2: <math>U_2U_1U_1R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>2 \cdot \frac{7!}{4!2!} = 210</math> possibilities for this case.<br />
<br />
Case 3: <math>U_2U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>\frac{6!}{2!2!} = 180</math> possibilities for this case.<br />
<br />
Case 4: <math>U_2U_2R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{6!}{2!4!} = 30</math> possibilities for this case.<br />
<br />
Case 5: <math>U_2U_2R_2R_1R_1</math> or <math>U_2U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{5!}{2!2!} = 60</math> possibilities for this case.<br />
<br />
Case 6: <math>U_2U_2R_2R_2</math><br />
<br />
There are <math>\frac{4!}{2!2!} = 6</math> possibilities for this case.<br />
<br />
Adding up all these cases gives us <math>70+210+180+30+60+6=\boxed{556}</math> ways.<br />
<br />
==Solution 3 (General Case)==<br />
<br />
Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these <math>4</math> points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath><br />
<br />
Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath><br />
<br />
<cmath>\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)</cmath><br />
<br />
<cmath>\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)</cmath><br />
<br />
<cmath>\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)</cmath><br />
<br />
<cmath>\varphi (1,1)=\varphi (1,0)+\varphi (0,1)</cmath><br />
<br />
Start with <math>\varphi (0,0)=1</math>, use this method and draw the figure below, we can finally get <cmath>\varphi (4,4)=556</cmath> (In order to make the LaTeX thing more beautiful to look at, I put <math>0</math> to make every number a <math>3</math>-digits number)<br />
<br />
<cmath>005-020-071-207-\boxed{556}</cmath> <cmath>003-010-032-084-207</cmath> <cmath>002-005-014-032-071</cmath> <cmath>001-002-005-010-020</cmath> <cmath>001-001-002-003-005</cmath><br />
<br />
So the total number of distinct sequences of jumps for the frog to reach <math>(4,4)</math> is <math>\boxed {556}</math>.<br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Video Solution==<br />
<br />
On The Spot STEM : <br />
https://www.youtube.com/watch?v=v2fo3CaAhmM&t=11s<br />
<br />
{{AIME box|year=2018|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}<br />
<br />
==Video Solution==<br />
On The Spot STEM Video: <br />
https://www.youtube.com/watch?v=v2fo3CaAhmM&t=11s</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_8&diff=1075082018 AIME II Problems/Problem 82019-07-09T01:45:15Z<p>Gamemaster402: /* Solution 3 (General Case) */</p>
<hr />
<div>==Problem==<br />
<br />
A frog is positioned at the origin of the coordinate plane. From the point <math>(x, y)</math>, the frog can jump to any of the points <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, or <math>(x, y + 2)</math>. Find the number of distinct sequences of jumps in which the frog begins at <math>(0, 0)</math> and ends at <math>(4, 4)</math>.<br />
<br />
==Solution 1==<br />
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to <math>(4,4)</math> in one move are <math>(2,4),(3,4),(4,2),</math> and <math>(4,3)</math>. This applies to any other point, thus we can work our way from <math>(0,0)</math> to <math>(4,4)</math>, recording down the number of ways to get to each point recursively. <br />
<br />
<math>(0,0): 1</math><br />
<br />
<math>(1,0)=(0,1)=1</math><br />
<br />
<math>(2,0)=(0, 2)=2</math><br />
<br />
<math>(3,0)=(0, 3)=3</math><br />
<br />
<math>(4,0)=(0, 4)=5</math><br />
<br />
<math>(1,1)=2</math>, <math>(1,2)=(2,1)=5</math>, <math>(1,3)=(3,1)=10</math>, <math>(1,4)=(4,1)= 20</math><br />
<br />
<math>(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71</math><br />
<br />
<math>(3,3)=84, (3,4)=(4,3)=207</math><br />
<br />
<math>(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}</math><br />
<br />
A diagram of the numbers:<br />
<br />
5 - 20 - 71 - 207 - <math>\boxed{556}</math><br />
<br />
3 - 10 - 32 - 84 - 207<br />
<br />
2 - 5 - 14 - 32 - 71<br />
<br />
1 - 2 - 5 - 10 - 20<br />
<br />
1 - 1 - 2 - 3 - 5<br />
<br />
==Solution 2==<br />
We'll refer to the moves <math>(x + 1, y)</math>, <math>(x + 2, y)</math>, <math>(x, y + 1)</math>, and <math>(x, y + 2)</math> as <math>R_1</math>, <math>R_2</math>, <math>U_1</math>, and <math>U_2</math>, respectively. Then the possible sequences of moves that will take the frog from <math>(0,0)</math> to <math>(4,4)</math> are all the permutations of <math>U_1U_1U_1U_1R_1R_1R_1R_1</math>, <math>U_2U_1U_1R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_1R_1</math>, <math>U_2U_2R_1R_1R_1R_1</math>, <math>U_1U_1U_1U_1R_2R_2</math>, <math>U_2U_2R_2R_1R_1</math>, <math>U_2U_1U_1R_2R_2</math>, and <math>U_2U_2R_2R_2</math>. We can reduce the number of cases using symmetry.<br />
<br />
Case 1: <math>U_1U_1U_1U_1R_1R_1R_1R_1</math><br />
<br />
There are <math>\frac{8!}{4!4!} = 70</math> possibilities for this case.<br />
<br />
Case 2: <math>U_2U_1U_1R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>2 \cdot \frac{7!}{4!2!} = 210</math> possibilities for this case.<br />
<br />
Case 3: <math>U_2U_1U_1R_2R_1R_1</math><br />
<br />
There are <math>\frac{6!}{2!2!} = 180</math> possibilities for this case.<br />
<br />
Case 4: <math>U_2U_2R_1R_1R_1R_1</math> or <math>U_1U_1U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{6!}{2!4!} = 30</math> possibilities for this case.<br />
<br />
Case 5: <math>U_2U_2R_2R_1R_1</math> or <math>U_2U_1U_1R_2R_2</math><br />
<br />
There are <math>2 \cdot \frac{5!}{2!2!} = 60</math> possibilities for this case.<br />
<br />
Case 6: <math>U_2U_2R_2R_2</math><br />
<br />
There are <math>\frac{4!}{2!2!} = 6</math> possibilities for this case.<br />
<br />
Adding up all these cases gives us <math>70+210+180+30+60+6=\boxed{556}</math> ways.<br />
<br />
==Solution 3 (General Case)==<br />
<br />
Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these <math>4</math> points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath><br />
<br />
Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath><br />
<br />
<cmath>\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)</cmath><br />
<br />
<cmath>\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)</cmath><br />
<br />
<cmath>\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)</cmath><br />
<br />
<cmath>\varphi (1,1)=\varphi (1,0)+\varphi (0,1)</cmath><br />
<br />
Start with <math>\varphi (0,0)=1</math>, use this method and draw the figure below, we can finally get <cmath>\varphi (4,4)=556</cmath> (In order to make the LaTeX thing more beautiful to look at, I put <math>0</math> to make every number a <math>3</math>-digits number)<br />
<br />
<cmath>005-020-071-207-\boxed{556}</cmath> <cmath>003-010-032-084-207</cmath> <cmath>002-005-014-032-071</cmath> <cmath>001-002-005-010-020</cmath> <cmath>001-001-002-003-005</cmath><br />
<br />
So the total number of distinct sequences of jumps for the frog to reach <math>(4,4)</math> is <math>\boxed {556}</math>.<br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
{{AIME box|year=2018|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}<br />
<br />
==Video Solution==<br />
On The Spot STEM Video: <br />
https://www.youtube.com/watch?v=v2fo3CaAhmM&t=11s</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_12&diff=1072022016 AIME II Problems/Problem 122019-07-01T05:37:46Z<p>Gamemaster402: /* Solution 8 */</p>
<hr />
<div>==Problem==<br />
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.<br />
<br />
<asy><br />
draw(Circle((0,0), 4));<br />
draw(Circle((0,0), 3));<br />
draw((0,4)--(0,3));<br />
draw((0,-4)--(0,-3));<br />
draw((-2.598, 1.5)--(-3.4641, 2));<br />
draw((-2.598, -1.5)--(-3.4641, -2));<br />
draw((2.598, -1.5)--(3.4641, -2));<br />
draw((2.598, 1.5)--(3.4641, 2));<br />
</asy><br />
<br />
==Solution 1==<br />
Choose a section to start coloring. Assume, WLOG, that this section is color <math>1</math>. We proceed coloring clockwise around the ring. Let <math>f(n,C)</math> be the number of ways to color the first <math>n</math> sections (proceeding clockwise) such that the last section has color <math>C</math>. In general (except for when we complete the coloring), we see that<br />
<cmath>f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),</cmath><br />
i.e., <math>f(n,C_i)</math> is equal to the number of colorings of <math>n-1</math> sections that end in any color other than <math>C_i</math>. Using this, we can compute the values of <math>f(n,C)</math> in the following table.<br />
<br />
<math>\begin{tabular}{c|c|c|c|c }<br />
\multicolumn{1}{c}{}&\multicolumn{4}{c}{\(C\)}\\<br />
\(n\)&1 & 2 & 3& 4 \\ \hline<br />
1& 1 & 0 & 0 & 0\\<br />
2 & 0 & 1 & 1 & 1 \\<br />
3& 3 & 2 & 2 & 2 \\<br />
4 & 6 & 7 & 7 & 7 \\<br />
5 & 21 & 20 & 20 & 20\\<br />
6& 0& 61 & 61 & 61\\<br />
\end{tabular}</math><br />
<br />
Note that <math>f(6, 1)=0</math> because then <math>2</math> adjacent sections are both color <math>1</math>. We multiply this by <math>4</math> to account for the fact that the initial section can be any color. Thus the desired answer is <math>(61+61+61) \cdot 4 = \boxed{732}</math>.<br />
<br />
Solution by Shaddoll<br />
<br />
==Solution 2==<br />
We use complementary counting. There are <math>4^6</math> total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of these as borders). Call these <math>B_1,B_2,\dots,B_6</math>. Let <math>\mathcal{A}_1, \mathcal{A}_2,\dots,\mathcal{A}_6</math> be the sets of colorings of the ring where the sections on both sides of <math>B_1,B_2,\dots,B_6</math> are the same color. We wish to determine <math>|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|</math>. Note that all of these cases are symmetric, and in general, <math>|\mathcal{A}_i|=4^5</math>. There are <math>6</math> such sets <math>\mathcal{A}_i</math>. Also, <math>|\mathcal{A}_i\cup\mathcal{A}_j|=4^4</math>, because we can only change colors at borders, so if we have two borders along which we cannot change colors, then there are four borders along which we have a choice of color. There are <math>\binom{6}{2}</math> such pairs <math>\mathcal{A}_i\cup\mathcal{A}_j</math>. Similarly, <math>|\mathcal{A}_i\cup \mathcal{A}_j\cup \mathcal{A}_k|=4^3</math>, with <math>\binom{6}{3}</math> such triples, and we see that the pattern will continue for 4-tuples and 5-tuples. For 6-tuples, however, these cases occur when there are no changes of color along the borders, i.e., each section has the same color. Clearly, there are four such possibilities.<br />
<br />
Therefore, by PIE,<br />
<cmath>|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|=\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4.</cmath><br />
We wish to find the complement of this, or <br />
<cmath>4^6-\left(\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4\right).</cmath><br />
By the Binomial Theorem, this is <math>(4-1)^6+3=\boxed{732}</math>.<br />
<br />
==Solution 3==<br />
We use generating functions. Suppose that the colors are <math>0,1,2,3</math>. Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors <math>s_i</math> and <math>s_{i+1}</math>, there exists a number <math>d_i\in\{1,2,3\}</math> such that <math>s_{i+1}\equiv s_i+d_i\pmod{4}</math>. Therefore, we can represent each border between sections by the generating function <math>(x+x^2+x^3)</math>, where <math>x,x^2,x^3</math> correspond to increasing the color number by <math>1,2,3\pmod4</math>, respectively. Thus the generating function that represents going through all six borders is <math>A(x)=(x+x^2+x^3)^6</math>, where the coefficient of <math>x^n</math> represents the total number of colorings where the color numbers are increased by a total of <math>n</math> as we proceed around the ring. But if we go through all six borders, we must return to the original section, which is already colored. Therefore, we wish to find the sum of the coefficients of <math>x^n</math> in <math>A(x)</math> with <math>n\equiv 0\pmod4</math>.<br />
<br />
Now we note that if <math>P(x)=x^n</math>, then<br />
<cmath>P(x)+P(ix)+P(-x)+P(-ix)=\begin{cases}4x^n&\text{if } n\equiv0\pmod{4}\\0&\text{otherwise}.\end{cases}</cmath><br />
Therefore, the sum of the coefficients of <math>A(x)</math> with powers congruent to <math>0\pmod 4</math> is<br />
<cmath>\frac{A(1)+A(i)+A(-1)+A(-i)}{4}=\frac{3^6+(-1)^6+(-1)^6+(-1)^6}{4}=\frac{732}{4}.</cmath><br />
We multiply this by <math>4</math> to account for the initial choice of color, so the answer is <math>\boxed{732}</math>.<br />
<br />
==Solution 4==<br />
Let <math>f(n)</math> be the number of valid ways to color a ring with <math>n</math> sections (which we call an <math>n</math>-ring), so the answer is given by <math>f(6)</math>. For <math>n=2</math>, we compute <math>f(n)=4\cdot3=12</math>. For <math>n \ge 3</math>, we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in <math>4</math> ways. Moving clockwise around the ring, there are <math>3</math> ways to color each of the <math>n-1</math> other sections. Therefore, we have <math>4 \cdot 3^{n-1}</math> colorings of an <math>n</math>-ring.<br />
<br />
However, note that the first and last sections could be the same color under this method. To count these invalid colorings, we see that by "merging" the first and last sections into one, we get a valid coloring of an <math>(n-1)</math>-ring. That is, there are <math>f(n-1)</math> colorings of an <math>n</math>-ring in which the first and last sections have the same color. Thus, <math>f(n) = 4 \cdot 3^{n-1} - f(n-1)</math> for all <math>n \ge 3</math>.<br />
<br />
To compute the requested value <math>f(6)</math>, we repeatedly apply this formula: <cmath>\begin{align*} f(6)&=4\cdot3^5-f(5)\\&=4\cdot3^5-4\cdot3^4+f(4)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-f(3)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-4\cdot3^2+f(2)\\&=4(3^5-3^4+3^3-3^2+3)\\&=4\cdot3\cdot\frac{3^5+1}{3+1}\\&=\boxed{732}.\end{align*}</cmath> (Solution by MSTang.)<br />
<br />
<br />
==Solution 5==<br />
Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colours of sections 1, 3, 5.<br />
<br />
Case 1: the colours of the three sections are the same.<br />
In this case, each of sections 2, 4, 6 can be one of 3 colours, so this case yields <math>4 \times 3^3 = 108</math> ways.<br />
<br />
Case 2: two of sections 1, 3, 5 are the same colour.<br />
Note that there are 3 ways for which two of the three sections have the same colour, and <math>4 \times 3 = 12</math> ways to determine their colours. After this, the section between the two with the same colour can be one of 3 colours and the other two can be one of 2 colours. This case yields <math>3 \times (4 \times 3) \times (3 \times 2 \times 2) = 432</math> ways.<br />
<br />
Case 3: all three sections of 1, 3, 5 are of different colours.<br />
Clearly there are <math>4 \times 3 \times 2 = 24</math> choices for which three colours to use, and there are 2 ways to choose the colours of each of sections 2, 4, 6. Thus, this case gives <math>4 \times 3 \times 2 \times 2^3 = 192</math> ways.<br />
<br />
In total, there are <math>108 + 432 + 192 = \boxed{732}</math> valid colourings.<br />
<br />
Solution by ADMathNoob<br />
<br />
<br />
==Solution 6==<br />
<br />
We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with <math>1, 2, </math> and <math>3</math> compartments, which are <math>4, 12, </math> and <math>24.</math> Now we will try to find a recursive formula, <math>C(n)</math>, for a circle with an arbitrary number of compartments <math>n.</math> We will do this by focusing on the <math>n-1</math> section in the circle. This section can either be the same color as the first compartment, or it can be a different color as the first compartment. We will focus on each case separately.<br />
<br />
Case 1:<br />
<br />
If they are the same color, we can say there are <math>C(n-2)</math> to fill the first <math>n-1</math> compartments. The <math>nth</math> compartment must be different from the first and second to last compartments, which are the same color. Hence this case adds <math>3*C(n-2)</math> to our recursive formula.<br />
<br />
Case 2:<br />
<br />
If they are different colors, we can say that there are <math>C(n-1)</math> to fill the first <math>n-1</math> compartments, and for the the <math>nth</math> compartment, there are <math>2</math> ways to color it because the <math>n-1</math> and <math>1</math> compartments are different colors. Hence this case adds <math>2*C(n-1).</math><br />
<br />
So our recursive formula, <math>C(n)</math>, is <math>3*C(n-2) + 2*C(n-1).</math> Using the initial values we calculated, we can evaluate this recursive formula up to <math>n=6.</math> When <math>n=6,</math> we get <math>\boxed{732}</math> valid colorings.<br />
<br />
Solution by NeeNeeMath<br />
<br />
==Solution 7==<br />
WLOG, color the top left section <math>1</math> and the top right section <math>2</math>. Then the left section can be colored <math>2</math>, <math>3</math>, or <math>4</math>, and the right section can be colored <math>1</math>, <math>3</math>, or <math>4</math>. There are <math>3 \cdot 3 = 9</math> ways to color the left and right sections. We split this up into two cases.<br />
<br />
Case 1: The left and right sections are of the same color. There are <math>2</math> ways this can happen: either they both are <math>3</math> or they both are <math>4</math>. We have <math>3</math> colors to choose for the bottom left, and <math>2</math> remaining colors to choose for the bottom right, for a total of <math>2 \cdot 3 \cdot 2</math> cases.<br />
<br />
Case 2: The left and right sections are of different colors. There are <math>9 - 2 = 7</math> ways this can happen. Assume the left is <math>3</math> and the right is <math>4</math>. Then the bottom left can be <math>1</math>, <math>2</math>, or <math>4</math>, and the bottom right can be <math>1</math>, <math>2</math>, or <math>3</math>. However the bottom sections cannot both be <math>1</math> or both be <math>2</math>, so there are <math>3 \cdot 3 - 2 = 7</math> ways to color the bottom sections, for a total for <math>7 \cdot 7 = 49</math> colorings.<br />
<br />
Since there were <math>4 \cdot 3 = 12</math> ways to color the top sections, the answer is <math>12 \cdot (12 + 49) = \boxed{732}</math>.<br />
<br />
==Solution 8==<br />
This is equivalent to a node coloring of a cycle with 6 nodes. After repeatedly using deletion-contraction, the solution comes out to be <math>\boxed{732}</math><br />
<br />
==Solution 9==<br />
Video Solution:<br />
https://www.youtube.com/watch?v=Yndl8HqVkJk<br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=II|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_14&diff=1069602018 AIME I Problems/Problem 142019-06-25T05:08:04Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div><br />
==Problem==<br />
<br />
<br />
<br />
Let <math>SP_1P_2P_3EP_4P_5</math> be a heptagon. A frog starts jumping at vertex <math>S</math>. From any vertex of the heptagon except <math>E</math>, the frog may jump to either of the two adjacent vertices. When it reaches vertex <math>E</math>, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than <math>12</math> jumps that end at <math>E</math>.<br />
<br />
==Solution==<br />
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line <math>E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E</math>. We can count the number of left/right (L/R) paths of length <math>\le 11</math> that start at <math>S</math> and end at either <math>P_4</math> or <math>P_3</math>.<br />
<br />
We can visualize the paths using the common grid counting method by starting at the origin <math>(0,0)</math>, so that a right (R) move corresponds to moving 1 in the positive <math>x</math> direction, and a left (L) move corresponds to moving 1 in the positive <math>y</math> direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines <math>y = x+2</math> or <math>y = x-3</math>. Letting <math>p(x,y)</math> be the number of such paths from <math>(0,0)</math> to <math>(x,y)</math> under these constraints, we have the following base cases:<br />
<br />
<math>p(x,0) = \begin{cases}<br />
1 & x \le 3 \\<br />
0 & x > 3<br />
\end{cases}<br />
\qquad p(0,y) = \begin{cases}<br />
1 & y \le 2 \\<br />
0 & y > 2<br />
\end{cases}<br />
</math><br />
<br />
and recursive step<br />
<math>p(x,y) = p(x-1,y) + p(x,y-1)</math> for <math>x,y \ge 1</math>.<br />
<br />
The filled in grid will look something like this, where the lower-left <math>1</math> corresponds to the origin:<br />
<br />
<math><br />
\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline<br />
0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline<br />
0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline<br />
0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline<br />
0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline<br />
\textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline<br />
1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline<br />
1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline<br />
\end{tabular}<br />
</math><br />
<br />
The bolded numbers on the top diagonal represent the number of paths from <math>S</math> to <math>P_4</math> in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from <math>S</math> to <math>P_3</math> in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line <math>x+y = 11</math>. The total number of ways is <math>1+3+9+28+89+1+4+14+47+155 = \boxed{351}</math>.<br />
<br />
(Solution by scrabbler94)<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=uWNExJc3hok<br />
<br />
==One page solution==<br />
<br />
[[Image:Aaaaaaaaaa.jpg|700px]]<br />
(Solution by mathleticguyyy)<br />
<br />
{{AIME box|year=2018|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}<br />
<br />
==Video Solution==<br />
To see a comprehensive video covering this, visit :<br />
https://www.youtube.com/watch?v=uWNExJc3hok</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_14&diff=1069592018 AIME I Problems/Problem 142019-06-25T05:07:14Z<p>Gamemaster402: /* One page solution */</p>
<hr />
<div><br />
==Problem==<br />
<br />
<br />
<br />
Let <math>SP_1P_2P_3EP_4P_5</math> be a heptagon. A frog starts jumping at vertex <math>S</math>. From any vertex of the heptagon except <math>E</math>, the frog may jump to either of the two adjacent vertices. When it reaches vertex <math>E</math>, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than <math>12</math> jumps that end at <math>E</math>.<br />
<br />
==Solution==<br />
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line <math>E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E</math>. We can count the number of left/right (L/R) paths of length <math>\le 11</math> that start at <math>S</math> and end at either <math>P_4</math> or <math>P_3</math>.<br />
<br />
We can visualize the paths using the common grid counting method by starting at the origin <math>(0,0)</math>, so that a right (R) move corresponds to moving 1 in the positive <math>x</math> direction, and a left (L) move corresponds to moving 1 in the positive <math>y</math> direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines <math>y = x+2</math> or <math>y = x-3</math>. Letting <math>p(x,y)</math> be the number of such paths from <math>(0,0)</math> to <math>(x,y)</math> under these constraints, we have the following base cases:<br />
<br />
<math>p(x,0) = \begin{cases}<br />
1 & x \le 3 \\<br />
0 & x > 3<br />
\end{cases}<br />
\qquad p(0,y) = \begin{cases}<br />
1 & y \le 2 \\<br />
0 & y > 2<br />
\end{cases}<br />
</math><br />
<br />
and recursive step<br />
<math>p(x,y) = p(x-1,y) + p(x,y-1)</math> for <math>x,y \ge 1</math>.<br />
<br />
The filled in grid will look something like this, where the lower-left <math>1</math> corresponds to the origin:<br />
<br />
<math><br />
\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline<br />
0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline<br />
0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline<br />
0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline<br />
0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline<br />
\textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline<br />
1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline<br />
1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline<br />
\end{tabular}<br />
</math><br />
<br />
The bolded numbers on the top diagonal represent the number of paths from <math>S</math> to <math>P_4</math> in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from <math>S</math> to <math>P_3</math> in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line <math>x+y = 11</math>. The total number of ways is <math>1+3+9+28+89+1+4+14+47+155 = \boxed{351}</math>.<br />
<br />
(Solution by scrabbler94)<br />
<br />
==One page solution==<br />
<br />
[[Image:Aaaaaaaaaa.jpg|700px]]<br />
(Solution by mathleticguyyy)<br />
<br />
{{AIME box|year=2018|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}<br />
<br />
==Video Solution==<br />
To see a comprehensive video covering this, visit :<br />
https://www.youtube.com/watch?v=uWNExJc3hok</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=1044612019 AIME I Problems/Problem 62019-03-15T05:43:41Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>==Problem 6==<br />
In convex quadrilateral <math>KLMN</math> side <math>\overline{MN}</math> is perpendicular to diagonal <math>\overline{KM}</math>, side <math>\overline{KL}</math> is perpendicular to diagonal <math>\overline{LN}</math>, <math>MN = 65</math>, and <math>KL = 28</math>. The line through <math>L</math> perpendicular to side <math>\overline{KN}</math> intersects diagonal <math>\overline{KM}</math> at <math>O</math> with <math>KO = 8</math>. Find <math>MO</math>.<br />
<br />
==Solution==<br />
Video Solution:<br />
https://www.youtube.com/watch?v=0AXF-5SsLc8<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_22&diff=1018992019 AMC 12A Problems/Problem 222019-02-12T05:48:28Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
<br />
Circles <math>\omega</math> and <math>\gamma</math>, both centered at <math>O</math>, have radii <math>20</math> and <math>17</math>, respectively. Equilateral triangle <math>ABC</math>, whose interior lies in the interior of <math>\omega</math> but in the exterior of <math>\gamma</math>, has vertex <math>A</math> on <math>\omega</math>, and the line containing side <math>\overline{BC}</math> is tangent to <math>\gamma</math>. Segments <math>\overline{AO}</math> and <math>\overline{BC}</math> intersect at <math>P</math>, and <math>\dfrac{BP}{CP} = 3</math>. Then <math>AB</math> can be written in the form <math>\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}</math> for positive integers <math>m</math>, <math>n</math>, <math>p</math>, <math>q</math> with <math>\gcd(m,n) = \gcd(p,q) = 1</math>. What is <math>m+n+p+q</math>?<br />
<math>\phantom{}</math><br />
<br />
<math>\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130</math><br />
<br />
==Solution==<br />
<br />
<asy><br />
size(20cm);<br />
draw(circle((0,0), 20));<br />
label("$\omega$", (0,0), 4.05*20*dir(149)*20/21);<br />
draw(circle((0,0), 17));<br />
label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21);<br />
dot((0,0));<br />
label("$O$", (0,0), E);<br />
pair aa = (-20, 0);<br />
dot(aa);<br />
label("$A$", aa, W);<br />
draw((-20,0)--(0,0));<br />
real a = (-20 + (80/sqrt(13) - 34/sqrt(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2));<br />
real ans = (80/sqrt(13) - 34/sqrt(3));<br />
dot((a,0));<br />
label("$P$", (a, 0), dir(290)*0.58);<br />
pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13);<br />
dot(s);<br />
label("$S$", s, dir(135));<br />
pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13));<br />
dot(c);<br />
label("$C$", c, dir(110));<br />
pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13));<br />
dot(m);<br />
label("$M$", m, dir(285));<br />
pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13));<br />
dot(b);<br />
label("$B$", b, S);<br />
draw(b--s);<br />
draw(s--(0,0));<br />
draw(aa--b);<br />
draw(aa--c);<br />
draw(aa--m);<br />
markscalefactor=0.1;<br />
draw(rightanglemark(s,m,aa,3.4));<br />
draw(rightanglemark((0,0),s,m,3.4));<br />
</asy><br />
<br />
Let <math>S</math> be the point of tangency between <math>\overline{BC}</math> and <math>\gamma</math>, and <math>M</math> be the midpoint of <math>\overline{BC}</math>. Note that <math>AM \perp BS</math> and <math>OS \perp BS</math>. This implies that <math>\angle OAM \cong \angle AOS</math>, and <math>\angle AMP \cong \angle OSP</math>. Thus, <math>\triangle PMA \sim \triangle PSO</math>.<br />
<br />
If we let <math>s</math> be the side length of <math>\triangle ABC</math>, then it follows that <math>AM = \frac{\sqrt{3}}{2}s</math> and <math>PM = \frac{s}{4}</math>. This implies that <math>AP = \frac{\sqrt{13}}{4}s</math>, so <math>\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}</math>. Furthermore, <math>\frac{AM + SO}{AO} = \frac{AM}{AP}</math> (because of <math>\triangle PMA \sim \triangle PSO</math>) so this gives us the equation<br />
<cmath>\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}</cmath><br />
to solve for the side length <math>s</math>, or <math>AB</math>. Thus,<br />
<cmath>\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}</cmath><br />
<cmath>\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}</cmath><br />
<cmath>s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB</cmath><br />
The problem asks for <math>m + n + p + q</math> which is <math>80 + 13 + 34 + 3</math> which is <math>130 = \boxed {E}</math>.<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=2eASfdhEyUE<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2019|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_22&diff=1018132017 AMC 12A Problems/Problem 222019-02-11T07:33:23Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
<br />
A square is drawn in the Cartesian coordinate plane with vertices at <math>(2, 2)</math>, <math>(-2, 2)</math>, <math>(-2, -2)</math>, <math>(2, -2)</math>. A particle starts at <math>(0,0)</math>. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is <math>1/8</math> that the particle will move from <math>(x, y)</math> to each of <math>(x, y + 1)</math>, <math>(x + 1, y + 1)</math>, <math>(x + 1, y)</math>, <math>(x + 1, y - 1)</math>, <math>(x, y - 1)</math>, <math>(x - 1, y - 1)</math>, <math>(x - 1, y)</math>, or <math>(x - 1, y + 1)</math>. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<br />
<math> \textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 39 </math><br />
<br />
==Solution==<br />
<br />
We let <math>c, e,</math> and <math>m</math> be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. <math>(1, 1)</math>), an "inside edge" (e.g. <math>(1, 0)</math>), and the middle respectively.<br />
<br />
Starting in the middle, there is a <math>\frac{4}{8}</math> chance of moving to an inside edge and a <math>\frac{4}{8}</math> chance of moving to an inside corner, so<br />
<br />
<cmath>m = \frac{1}{2}e + \frac{1}{2}c.</cmath><br />
<br />
Starting at an inside edge, there is a <math>\frac{2}{8}</math> chance of moving to another inside edge, a <math>\frac{2}{8}</math> chance of moving to an inside corner, a <math>\frac{1}{8}</math> chance of moving into the middle, and a <math>\frac{3}{8}</math> chance of reaching an outside edge and stopping. Therefore,<br />
<br />
<cmath>e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}*0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.</cmath><br />
<br />
Starting at an inside corner, there is a <math>\frac{2}{8}</math> chance of moving to an inside edge, a <math>\frac{1}{8}</math> chance of moving into the middle, a <math>\frac{4}{8}</math> chance of moving to an outside edge and stopping, and finally a <math>\frac{1}{8}</math> chance of reaching that elusive outside corner. This gives<br />
<br />
<cmath>c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}*1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.</cmath><br />
<br />
Solving this system of equations gives<br />
<br />
<cmath>m = \frac{4}{35},</cmath><br />
<cmath>e = \frac{1}{14},</cmath><br />
<cmath>c = \frac{11}{70}.</cmath><br />
<br />
Since the particle starts at <math>(0, 0),</math> it is <math>m</math> we are looking for, so the final answer is<br />
<br />
<cmath>4 + 35 = \boxed{\textbf{(E) }39}.</cmath><br />
==Video Solution==<br />
https://www.youtube.com/watch?v=rz-Ma_O2bT4<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_25&diff=1018122014 AMC 10B Problems/Problem 252019-02-11T07:32:20Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
In a small pond there are eleven lily pads in a row labeled <math>0</math> through <math>10</math>. A frog is sitting on pad <math>1</math>. When the frog is on pad <math>N</math>, <math>0<N<10</math>, it will jump to pad <math>N-1</math> with probability <math>\frac{N}{10}</math> and to pad <math>N+1</math> with probability <math>1-\frac{N}{10}</math>. Each jump is independent of the previous jumps. If the frog reaches pad <math>0</math> it will be eaten by a patiently waiting snake. If the frog reaches pad <math>10</math> it will exit the pond, never to return. What is the probability that the frog will escape before being eaten by the snake?<br />
<br />
<math> \textbf {(A) } \frac{32}{79} \qquad \textbf {(B) } \frac{161}{384} \qquad \textbf {(C) } \frac{63}{146} \qquad \textbf {(D) } \frac{7}{16} \qquad \textbf {(E) } \frac{1}{2} </math><br />
<br />
==Solution==<br />
<br />
Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math>P_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of <math>5</math> equations:<br />
<cmath>P_1=\frac{9}{10}\cdot P_2</cmath><br />
<cmath>P_2=\frac{2}{10}\cdot P_1 + \frac{8}{10}\cdot P_3</cmath><br />
<cmath>P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4</cmath><br />
<cmath>P_4=\frac{4}{10}\cdot P_3 + \frac{6}{10}\cdot P_5</cmath><br />
<cmath>P_5=\frac{5}{10}</cmath><br />
We want to find <math>P_1</math>, since the frog starts at pad <math>1</math>. Solving the above system (really long process) yields <math>P_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(C)}</math>.<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=DMdgh2mMiWM<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_25&diff=1018112018 AMC 10B Problems/Problem 252019-02-11T07:31:05Z<p>Gamemaster402: </p>
<hr />
<div>== Problem ==<br />
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?<br />
<br />
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math><br />
<br />
==Solution 1==<br />
This rewrites itself to <math>x^2=10,000\{x\}</math>.<br />
<br />
Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a hole at <math>x=2</math> etc.<br />
<br />
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.<br />
<br />
<asy><br />
import graph;<br />
size(400);<br />
xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));<br />
yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));<br />
real y(real x) {return x^2;}<br />
draw(circle((-4,16), 0.1));<br />
draw(circle((-3,16), 0.1));<br />
draw(circle((-2,16), 0.1));<br />
draw(circle((-1,16), 0.1));<br />
draw(circle((0,16), 0.1));<br />
draw(circle((1,16), 0.1));<br />
draw(circle((2,16), 0.1));<br />
draw(circle((3,16), 0.1));<br />
draw(circle((4,16), 0.1));<br />
draw((-5,0)--(-4,16), black);<br />
draw((-4,0)--(-3,16), black);<br />
draw((-3,0)--(-2,16), black);<br />
draw((-2,0)--(-1,16), black);<br />
draw((-1,0)--(-0,16), black);<br />
draw((0,0)--(1,16), black);<br />
draw((1,0)--(2,16), black);<br />
draw((2,0)--(3,16), black);<br />
draw((3,0)--(4,16), black);<br />
draw(graph(y,-4.2,4.2),green);<br />
</asy><br />
<br />
Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>.<br />
<br />
Note: from the graph we can clearly see there are 4 solution on the negative side and only 2 on the positive side. So the solution really should be from -100 to 98, which still counts to 199. A couple of the alternative solutions also seem to have the same flaw.<br />
<br />
==Solution 2==<br />
<br />
Same as the first solution, <math>x^2=10,000\{x\} </math>.<br />
<br />
<br />
We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>{x}</math> in terms of <math>\lfloor x \rfloor</math>:<br />
<math>\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</math><br />
<br />
<br />
We use the quadratic formula to solve for {x}:<br />
<math>\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2} </math><br />
<br />
<br />
Since <math> 0 \leq \{x\} < 1 </math>, we get an inequality which we can then solve. After simplifying a lot, we get that <math>\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0</math>.<br />
<br />
<br />
Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.<br />
<br />
==Solution 3==<br />
<br />
Let <math>x = a+k</math> where <math>a</math> is the integer part of <math>x</math> and <math>k</math> is the fractional part of <math>x</math>.<br />
We can then rewrite the problem below:<br />
<br />
<math>(a+k)^2 + 10000a = 10000(a+k)</math><br />
<br />
From here, we get<br />
<br />
<math>(a+k)^2 + 10000a = 10000a + 10000k</math><br />
<br />
Solving for <math>a+k = x</math><br />
<br />
<math>(a+k)^2 = 10000k</math><br />
<br />
<math>x = a+k = \pm100\sqrt{k}</math><br />
<br />
Because <math>0 \leq k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore:<br />
<br />
<math>-99 \leq x \leq 99</math><br />
<br />
There are 199 elements in this range, so the answer is <math>\boxed{\textbf{(C)} \text{ 199}}</math>.<br />
<br />
==Solution 4==<br />
<br />
Notice the given equation is equivilent to <math>(\lfloor x \rfloor+\{x\})^2=10,000\{x\} </math><br />
<br />
Now we now that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values.<br />
<br />
<math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math><br />
<br />
<math>(\lfloor x \rfloor +1) = \pm 100</math><br />
<br />
<math>\lfloor x \rfloor = 99, -101</math><br />
<br />
And just like Solution 2, we see that <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.<br />
<br />
==Solution 5==<br />
<br />
First, we can let <math>\{x\} = b, \lfloor x \rfloor = a</math>. We know that <math>a + b = x</math> by definition. We can rearrange the equation to obtain <br />
<br />
<math>x^2 = 10^4(x - a)</math>. <br />
<br />
By taking square root on both sides, we obtain <math>x = \pm 100 \sqrt{b}</math> (because <math>x - a = b</math>). We know since <math>b</math> is the fractional part of <math>x</math>, it must be that <math>0 \leq b < 1</math>. Thus, <math>x</math> may take any value in the interval <math>-100 < x < 100</math>. Hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values for <math>\lfloor x \rfloor</math> in that range and we are done. <br />
<br />
~awesome1st<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=vHKPbaXwJUE<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=1018102018 AMC 10B Problems/Problem 242019-02-11T07:30:09Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\frac{\sqrt{3}}{4}</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}}{16}</math> for a total area of <math>\frac{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\frac{\sqrt{3}}{4}</math> giving us an area of <math>\frac{\sqrt{3}}{32}</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}</math>. Taking the complement, we get <math>\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==Solution 4==<br />
Dividing <math>\triangle MNO</math> into two right triangles congruent to <math>\triangle PMN</math>, we see that <math>[MPNQOR]=\dfrac{5}{8}[ACE]</math>. Because <math>[ACE] = \dfrac{1}{2}[ABCDEF]</math>, we have <math>[MPNQOR]=\dfrac{5}{16}[ABCDEF]</math>. From here, you should be able to tell that the answer will have a factor of <math>5</math>, and <math>\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}</math> is the only answer that has a factor of <math>5</math>. However, if you want to actually calculate the area, you would calculate <math>[ABCDEF]</math> to be <math>6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}</math>, so <math>[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}</math>.<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=yDbn9Mx2myw<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1018092018 AMC 10B Problems/Problem 232019-02-11T07:29:10Z<p>Gamemaster402: </p>
<hr />
<div>How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
<br />
<br />
==Solution==<br />
Let <math>x =</math> lcm<math>(a, b)</math>, and <math>y = </math>gcd<math>(a, b)</math>. Therefore, <math>a\cdot b = </math>lcm<math>(a, b)\cdot </math>gcd<math>(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
<br />
Using Simon's Favorite Factoring Trick, we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
<br />
Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
<br />
Edited by IronicNinja~<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=JWGHYUeOx-k<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_22&diff=1018082018 AMC 10B Problems/Problem 222019-02-11T07:27:58Z<p>Gamemaster402: </p>
<hr />
<div>Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>[0,1]</math>. Which of the following numbers is closest to the probability that <math>x,y,</math> and <math>1</math> are the side lengths of an obtuse triangle?<br />
<br />
<math>\textbf{(A)} \text{ 0.21} \qquad \textbf{(B)} \text{ 0.25} \qquad \textbf{(C)} \text{ 0.29} \qquad \textbf{(D)} \text{ 0.50} \qquad \textbf{(E)} \text{ 0.79}</math><br />
<br />
== Solution == <br />
The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. So, we have two inequalities:<br />
<cmath>x^2 + y^2 < 1</cmath><br />
<cmath>x + y > 1</cmath><br />
The first equation is <math>\frac14</math> of a circle with radius <math>1</math>, and the second equation is a line from <math>(0, 1)</math> to <math>(1, 0)</math>.<br />
So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>0.29</math>.<br />
<br />
==Solution 2==<br />
<br />
Note that the obtuse angle in the triangle has to be opposite the side that is always length 1. This is because the largest angle is always opposite the largest side, and if 2 sides of the triangle were 1, the last side would have to be greater than 1 to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite 1:<br />
<br />
<cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath><br />
<br />
where <math>x</math> and <math>y</math> are the sides that go from <math>[0,1]</math> and <math>\theta</math> is the angle opposite the side of length 1.<br />
<br />
By isolating <math>\cos(\theta)</math>, we get:<br />
<br />
<cmath>\frac{1-x^2-y^2}{-2xy} = \cos(\theta)</cmath><br />
<br />
For <math>\theta</math> to be obtuse, <math>\cos(\theta)</math> must be negative. Therefore, <math>\frac{1-x^2-y^2}{-2xy}</math> is negative. Since <math>x</math> and <math>y</math> must be positive, <math>-2xy</math> must be negative, so we must make <math>1-x^2-y^2</math> positive. From here, we can set up the inequality<br />
<cmath>x^2+y^2<1</cmath><br />
Additionally, to satisfy the definition of a triangle, we need:<br />
<cmath>x+y>1</cmath><br />
The solution should be the overlap between the two equations in the 1st quadrant.<br />
<br />
By observing that <math>x^2+y^2<1</math> is the equation for a circle, the amount that is in the 1st quadrant is <math>\frac{\pi}{4}</math>. The line can also be seen as a chord that goes from <math>(0, 1)</math> to <math>(1, 0)</math>. By cutting off the triangle of area <math>\frac{1}{2}</math> that is not part of the overlap, we get <math>\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}</math>.<br />
<br />
-allenle873<br />
==Solution through video==<br />
https://www.youtube.com/watch?v=GHAMU60rI5c<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=1018072018 AMC 10B Problems/Problem 212019-02-11T07:25:58Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>?<br />
<br />
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math><br />
<br />
==Solution 1==<br />
<br />
Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. The desired answer needs to be a multiple of <math>17</math> or <math>19</math>, because if it is not a multiple of <math>17</math> or <math>19</math>, the LCM, or the least possible value for <math>n</math>, will not be more than 4 digits. Looking at the answer choices, <math>\fbox{(C) 340}</math> is the smallest number divisible by <math>17</math> or <math>19</math>. Checking, we can see that <math>n</math> would be <math>6460</math>.<br />
<br />
==Solution 2==<br />
Let the next largest divisor be <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the gcd can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{(C) }340}</math>.<br />
<br />
==Video==<br />
https://www.youtube.com/watch?v=KHaLXNAkDWE<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=20|num-a=22}}<br />
{{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_13&diff=1018062016 AIME I Problems/Problem 132019-02-11T07:24:08Z<p>Gamemaster402: </p>
<hr />
<div>==Problem==<br />
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line <math>y = 24</math>. A fence is located at the horizontal line <math>y = 0</math>. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where <math>y=0</math>, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where <math>y < 0</math>. Freddy starts his search at the point <math>(0, 21)</math> and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.<br />
==Solution==<br />
Clearly Freddy's <math>x</math>-coordinate is irrelevant, so we let <math>E(y)</math> be the expected value of the number of jumps it will take him to reach the river from a given <math>y</math>-coordinate. Observe that <math>E(24)=0</math>, and <cmath>E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}</cmath> for all <math>y</math> such that <math>1\le y\le 23</math>. Also note that <math>E(0)=1+\frac{2E(0)+E(1)}{3}</math>. This gives <math>E(0)=E(1)+3</math>. Plugging this into the equation for <math>E(1)</math> gives that <cmath>E(1)=1+\frac{E(2)+3E(1)+3}{4},</cmath> or <math>E(1)=E(2)+7</math>. Iteratively plugging this in gives that <math>E(n)=E(n+1)+4n+3</math>. Thus <math>E(23)=E(24)+95</math>, <math>E(22)=E(23)+91=E(24)+186</math>, and <math>E(21)=E(22)+87=E(24)+273=\boxed{273}</math>.<br />
==Video Solution==<br />
For those who want more explanation, here is a video explaining the solution: https://www.youtube.com/watch?v=jQCGkOGMlFQ<br />
== See also ==<br />
{{AIME box|year=2016|n=I|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_15&diff=999602008 AIME II Problems/Problem 152019-01-02T06:19:04Z<p>Gamemaster402: /* Solution 3 (Big Bash) */</p>
<hr />
<div>== Problem ==<br />
Find the largest integer <math>n</math> satisfying the following conditions:<br />
:(i) <math>n^2</math> can be expressed as the difference of two consecutive cubes;<br />
:(ii) <math>2n + 79</math> is a perfect square.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Write <math>n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1</math>, or equivalently, <math>(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2</math>.<br />
<br />
Since <math>2n + 1</math> and <math>2n - 1</math> are both odd and their difference is <math>2</math>, they are [[relatively prime]]. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have <math>2n - 1</math> be three times a square, for then <math>2n + 1</math> would be a square congruent to <math>2</math> modulo <math>3</math>, which is impossible.<br />
<br />
Thus <math>2n - 1</math> is a square, say <math>b^2</math>. But <math>2n + 79</math> is also a square, say <math>a^2</math>. Then <math>(a + b)(a - b) = a^2 - b^2 = 80</math>. Since <math>a + b</math> and <math>a - b</math> have the same parity and their product is even, they are both even. To maximize <math>n</math>, it suffices to maximize <math>2b = (a + b) - (a - b)</math> and check that this yields an integral value for <math>m</math>. This occurs when <math>a + b = 40</math> and <math>a - b = 2</math>, that is, when <math>a = 21</math> and <math>b = 19</math>. This yields <math>n = 181</math> and <math>m = 104</math>, so the answer is <math>\boxed{181}</math>.<br />
<br />
=== Solution 2 ===<br />
Suppose that the consecutive cubes are <math>m</math> and <math>m + 1</math>. We can use completing the square and the first condition to get:<br />
<cmath><br />
(2n)^2 - 3(2m + 1)^2 = 1\equiv a^2 - 3b^2<br />
</cmath><br />
where <math>a</math> and <math>b</math> are non-negative integers. Now this is a [[Pell equation]], with solutions in the form <math>(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,</math> <math>k = 0,1,2,3,...</math>. However, <math>a</math> is even and <math>b</math> is odd. It is easy to see that the parity of <math>a</math> and <math>b</math> switch each time (by induction). Hence all solutions to the first condition are in the form:<br />
<cmath><br />
(2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k<br />
</cmath><br />
where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =\boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done.<br />
<br />
=== Solution 3 (BIG BAYUS) ===<br />
<br />
Let us generate numbers <math>1</math> to <math>1000</math> for the second condition, for squares. We know for <math>N</math> to be integer, the squares must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a</math> be the smaller consecutive cube. We then get:<br />
<br />
<math> N^2 = (A + 1)^3 - A^3</math><br />
<br />
<math> N^2 - 1 = 3A^2 + 3A</math><br />
<br />
<math> (N + 1)(N - 1) = 3A(A + 1)</math><br />
<br />
Now we know either <math>N + 1</math> or <math>N - 1</math> must be factor of <math>3</math>, hence <math>N = 1 (\mod 3) </math> or<math> N = 2 (\mod 3)</math>. Only <math>1, 73, 181, 721</math> satisfy this criterion. Testing each of the numbers in the condition yields <math>181</math> as the largest that fits both, thus answer <math>= \boxed{181}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|num-b=14|after=Last problem}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_14&diff=981921985 AIME Problems/Problem 142018-10-19T04:05:28Z<p>Gamemaster402: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded <math>1</math> point, the loser got <math>0</math> points, and each of the two players earned <math>\frac{1}{2}</math> point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?<br />
<br />
== Solution 1==<br />
Let us suppose for convenience that there were <math>n + 10</math> players overall. Among the <math>n</math> players not in the weakest 10 there were <math>n \choose 2</math> games played and thus <math>n \choose 2</math> points earned. By the givens, this means that these <math>n</math> players also earned <math>n \choose 2</math> points against our weakest 10. Now, the 10 weakest players playing amongst themselves played <math>{10 \choose 2} = 45</math> games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger <math>n</math> players. Since every point earned falls into one of these categories, It follows that the total number of points earned was <math>2{n \choose 2} + 90 = n^2 - n + 90</math>. However, there was one point earned per game, and there were a total of <math>{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}</math> games played and thus <math>\frac{(n + 10)(n + 9)}{2}</math> points earned. So we have <math>n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}</math> so <math>2n^2 - 2n + 180 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> players got <math>n(n - 1)</math> points in total (by our previous calculation) for an average of <math>n - 1</math>, while the bottom 10 got 90 points total, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>.<br />
== Solution 2 == <br />
Suppose that there are <math>n</math> players participating in the tournament. We break this up into a group of the weakest ten, and the other <math>n-10</math> people. Note that the <math>10</math> players who played each other generated a total of <math>\dbinom{10}{2} = 45</math> points playing each other. Thus, they earned <math>45</math> playing the <math>n-10</math> other people. Thus, the <math>n-10</math> people earned a total of <math>10(n-10)-45 = 10n-145</math> points playing vs. this group of 10 people, and also earned a total of <math>10n-145</math> playing against themselves. Since each match gives a total of one point, we must have that <math>\dbinom{n-10}{2}=10n-45</math>. Expanding and simplifying gives us <math>n^2-41+400=0</math>. Thus, <math>n=16</math> or <math>n=25</math>. Note however that if <math>n=16</math>, then the strongest <math>16</math> people get a total of <math>16*10-145=15</math> playing against the weakest <math>10</math> who gained <math>45</math> points vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.<br />
<br />
Solution by GameMaster402<br />
<br />
== See also ==<br />
{{AIME box|year=1985|num-b=13|num-a=15}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_14&diff=981911985 AIME Problems/Problem 142018-10-19T04:04:50Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded <math>1</math> point, the loser got <math>0</math> points, and each of the two players earned <math>\frac{1}{2}</math> point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?<br />
<br />
== Solution 1==<br />
Let us suppose for convenience that there were <math>n + 10</math> players overall. Among the <math>n</math> players not in the weakest 10 there were <math>n \choose 2</math> games played and thus <math>n \choose 2</math> points earned. By the givens, this means that these <math>n</math> players also earned <math>n \choose 2</math> points against our weakest 10. Now, the 10 weakest players playing amongst themselves played <math>{10 \choose 2} = 45</math> games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger <math>n</math> players. Since every point earned falls into one of these categories, It follows that the total number of points earned was <math>2{n \choose 2} + 90 = n^2 - n + 90</math>. However, there was one point earned per game, and there were a total of <math>{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}</math> games played and thus <math>\frac{(n + 10)(n + 9)}{2}</math> points earned. So we have <math>n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}</math> so <math>2n^2 - 2n + 180 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> players got <math>n(n - 1)</math> points in total (by our previous calculation) for an average of <math>n - 1</math>, while the bottom 10 got 90 points total, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>.<br />
== Solution 2 == <br />
Suppose that there are <math>n-10</math> players participating in the tournament. We break this up into a group of the weakest ten, and the other <math>n-10</math> people. Note that the <math>10</math> players who played each other generated a total of <math>\dbinom{10}{2} = 45</math> points playing each other. Thus, they earned <math>45</math> playing the <math>n-10</math> other people. Thus, the <math>n-10</math> people earned a total of <math>10(n-10)-45 = 10n-145</math> points playing vs. this group of 10 people, and also earned a total of <math>10n-145</math> playing against themselves. Since each match gives a total of one point, we must have that <math>\dbinom{n-10}{2}=10n-45</math>. Expanding and simplifying gives us <math>n^2-41+400=0</math>. Thus, <math>n=16</math> or <math>n=25</math>. Note however that if <math>n=16</math>, then the strongest <math>16</math> people get a total of <math>16*10-145=15</math> playing against the weakest <math>10</math> who gained <math>45</math> points vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.<br />
<br />
Solution by GameMaster402<br />
<br />
== See also ==<br />
{{AIME box|year=1985|num-b=13|num-a=15}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_2&diff=932051995 IMO Problems/Problem 22018-03-13T04:23:42Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
(''Nazar Agakhanov, Russia'')<br />
Let <math>a, b, c</math> be positive real numbers such that <math>abc = 1</math>. Prove that<br />
<cmath> \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. </cmath><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
We make the substitution <math>x= 1/a</math>, <math>y=1/b</math>, <math>z=1/c</math>. Then<br />
<cmath> \begin{align*}<br />
\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} &= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\<br />
&= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} .<br />
\end{align*} </cmath><br />
Since <math>(x^2,y^2,z^2)</math> and <math>\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \bigr)</math> are similarly sorted sequences, it follows from the [[Rearrangement Inequality]] that<br />
<cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{2} \left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) . </cmath><br />
By the [[Power Mean Inequality]],<br />
<cmath> \frac{y^2+z^2}{y+z} \ge \frac{(y+z)^2}{2(x+y)} = \frac{x+y}{2} . </cmath><br />
Symmetric application of this argument yields<br />
<cmath> \frac{1}{2}\left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) \ge \frac{1}{2}(x+y+z) . </cmath><br />
Finally, [[AM-GM]] gives us<br />
<cmath> \frac{1}{2}(x+y+z) \ge \frac{3}{2}xyz = \frac{3}{2}, </cmath><br />
as desired. <math>\blacksquare</math><br />
<br />
=== Solution 2 ===<br />
<br />
We make the same substitution as in the first solution. We note that in general,<br />
<cmath> \frac{p}{q+r} = \frac{(p+q+r)}{(p+q+r)-p} - 1 . </cmath><br />
It follows that <math>(x,y,z)</math> and <math>\bigl(x/(y+z), y/(z+x), z/(x+y)\bigr)</math> are similarly sorted sequences. Then by [[Chebyshev's Inequality]],<br />
<cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{3}(x+y+z) \left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \right) . </cmath><br />
By AM-GM, <math>\frac{x+y+z}{3} \ge \sqrt[3]{xyz}=1</math>, and by [[Nesbitt's Inequality]],<br />
<cmath> \frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \ge \frac{3}{2}. </cmath><br />
The desired conclusion follows. <math>\blacksquare</math><br />
<br />
=== Solution 3 ===<br />
Without clever substitutions:<br />
By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM.<br />
<br />
=== Solution 3b ===<br />
Without clever notation:<br />
By Cauchy-Schwarz, <cmath>\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)</cmath><br />
<cmath>\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2</cmath><br />
<cmath>= (ab + bc + ac)^2</cmath><br />
<br />
Dividing by <math>2(ab + bc + ac)</math> and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives<br />
<cmath>\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath><br />
as desired.<br />
<br />
=== Solution 4 ===<br />
Proceed as in Solution 1, to arrive at the equivalent inequality<br />
<cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath><br />
But we know that <cmath>x + y + z \ge 3xyz \ge 3</cmath> by AM-GM. Furthermore,<br />
<cmath> (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 </cmath><br />
by Cauchy-Schwarz, and so dividing by <math>2(x + y + z)</math> gives<br />
<cmath> \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &\ge \frac{(x + y + z)}{2} \\ &\ge \frac{3}{2} \end{align*},</cmath><br />
as desired.<br />
<br />
=== Solution 5 ===<br />
Without clever substitutions, and only AM-GM!<br />
<br />
Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math><br />
<br />
<br />
=== Solution 6 from Brilliant Wiki (Muirheads) ====<br />
https://brilliant.org/wiki/muirhead-inequality/<br />
<br />
Scroll all the way down<br />
{{alternate solutions}}<br />
<br />
== Resources ==<br />
<br />
* [[1995 IMO Problems]]<br />
* [http://www.mathlinks.ro/Forum/viewtopic.php?p=365178#p365178 Discussion on AoPS/MathLinks]<br />
<br />
<br />
[[Category:Olympiad Algebra Problems]]</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91949AMC historical results2018-02-18T20:43:10Z<p>Gamemaster402: /* AMC 12A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average Score: 61.77<br />
*AIME floor: 114.0<br />
*Distinguished Honor Roll: 130.5<br />
<br />
===AMC 10B===<br />
*Average score:<br />
*AIME floor:<br />
*Distinguished Honor Roll:<br />
<br />
===AMC 12A===<br />
*Average score: 58.97<br />
*AIME floor: 100.5<br />
*Distinguished Honor Roll floor: 121.5<br />
<br />
===AMC 12B===<br />
*Average score:<br />
*AIME floor:<br />
*Distinguished Honor Roll:<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91948AMC historical results2018-02-18T20:42:44Z<p>Gamemaster402: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average Score: 61.77<br />
*AIME floor: 114.0<br />
*Distinguished Honor Roll: 130.5<br />
<br />
===AMC 10B===<br />
*Average score:<br />
*AIME floor:<br />
*Distinguished Honor Roll:<br />
<br />
===AMC 12A===<br />
*Average score:<br />
*AIME floor:<br />
*Distinguished Honor Roll floor:<br />
<br />
===AMC 12B===<br />
*Average score:<br />
*AIME floor:<br />
*Distinguished Honor Roll:<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91131AMC historical results2018-02-12T23:15:30Z<p>Gamemaster402: /* AMC 12A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: 63.5<br />
*AIME floor: 114<br />
*Distinguished Honor Roll: 130.5<br />
<br />
===AMC 10B===<br />
This exam has not occurred yet.<br />
<br />
===AMC 12A===<br />
*Average score: 61.2<br />
*AIME floor: 100.5<br />
*Distinguished Honor Roll floor: 124.5<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91130AMC historical results2018-02-12T23:15:07Z<p>Gamemaster402: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: 63.5<br />
*AIME floor: 114<br />
*Distinguished Honor Roll: 130.5<br />
<br />
===AMC 10B===<br />
This exam has not occurred yet.<br />
<br />
===AMC 12A===<br />
*Average score: <br />
*AIME floor: <br />
*Distinguished Honor Roll floor:<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91056AMC historical results2018-02-11T22:48:00Z<p>Gamemaster402: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: 57.3<br />
*AIME floor: 120<br />
*Distinguished Honor Roll floor: 135.0<br />
<br />
===AMC 10B===<br />
This exam has not occurred yet.<br />
<br />
===AMC 12A===<br />
*Average score: <br />
*AIME floor: <br />
*Distinguished Honor Roll floor:<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_20&diff=897992012 AMC 10A Problems/Problem 202018-01-15T20:26:07Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #15]] and [[2012 AMC 10A Problems|2012 AMC 10A #20]]}}<br />
<br />
==Problem==<br />
<br />
A <math>3 \times 3</math> square is partitioned into <math>9</math> unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated <math>90\,^{\circ}</math> clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?<br />
<br />
<math> \textbf{(A)}\ \frac{49}{512}\qquad\textbf{(B)}\ \frac{7}{64}\qquad\textbf{(C)}\ \frac{121}{1024}\qquad\textbf{(D)}\ \frac{81}{512}\qquad\textbf{(E)}\ \frac{9}{32} </math><br />
<br />
== Solution 1==<br />
First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is <math>1</math> case with all black squares. There are four cases with one white square and all <math>4</math> work. There are six cases with two white squares, but only the <math>2</math> with the white squares diagonal from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is <math>1+4+2=7</math>. In essence, the edges work the same way, so there are also <math>7</math> ways to color them. The number of ways to fit the conditions over the number of ways to color the squares is<br />
<br />
<cmath>\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}</cmath><br />
<br />
<br />
==Solution 2==<br />
We proceed by casework.<br />
Note that the middle square must be black, because when rotated 90 degrees, it must keep its position. Now we have to deal with the following cases: <br />
Case 1: 0 white squares.<br />
There is exactly <math> 1 </math> way to color the grid this way.<br />
Case 2: 1 white square.<br />
Note that the white square can be anywhere on the grid except for the middle square, because when rotating 90 degrees it can never land on it self. Thus, there are <math>8</math> cases.<br />
Case 3: 2 white squares. <br />
We have <math>\binom{8}{2}=28</math> ways to color two white squares without restrictions (the middle square must be black, giving us 8 squares to choose from). However, we must subtract the ways in which two white squares differ by a rotation of 90 degrees about the middle of the square. There are a total of 8 cases we must subtract (these are not too hard to see). Thus, there are <math>20</math> ways from this.<br />
Case 4: 3 white squares.<br />
Since we can not change the middle square, there are <math>binom{8}{3}=56</math> ways to color this. However, we must consider the cases where at least two squares differ by a rotation of 90 degrees. We can count this with PIE: by the Principle of Exclusion, the number of cases we want to exclude is the number of cases where 2 squares differ by a rotation of 90 degrees and minus when there are 3 squares such that two of them differ by rotation of 90 and 1 of them differs by rotation of 180, because of the overcount from our first case. <br />
From case 2, there are 8 causes such that two squares differ by a rotation of 90. There are also 6 other places we can place the third square (it can't be the middle of the two that we already colored), for a total of 48 ways. We have to subtract the second case. Note that there are 8 ways in which we can arrange two squares differing by 180 degrees. Out of these, each one has two ways to put another square such that two differ by 90 degrees and 1 pair differs by 180. However, this is overcounted by a factor of 2, so there are actually 8*2/2=8 ways. <br />
Thus, we have <math>56-(48-8)=16</math> ways in this case.<br />
Case 5: 4 white squares.<br />
Note that two of them have to be on one of the 4 corner squares, and two of them have to be on one of the 4 edge squares. Each solution yields two combinations, for a total of 2*2=4.<br />
Adding up our cases yields <math>1+8+20+16+4=49</math> ways. There are 512 ways to color the square without restrictions. Thus, the answer is $\boxed{49/512}<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2012|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_8&diff=897152006 AIME II Problems/Problem 82018-01-13T04:47:11Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>== Problem ==<br />
There is an unlimited supply of [[congruent]] [[equilateral triangle]]s made of colored paper. Each [[triangle]] is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.<br />
<br />
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?<br />
<br />
<asy><br />
pair A,B;<br />
A=(0,0); B=(2,0);<br />
pair C=rotate(60,A)*B;<br />
pair D, E, F;<br />
D = (1,0);<br />
E=rotate(60,A)*D;<br />
F=rotate(60,C)*E;<br />
draw(C--A--B--cycle); draw(D--E--F--cycle);<br />
</asy><br />
<br />
== Solution 1==<br />
If two of our big equilateral triangles have the same color for their center triangle and the same [[multiset]] of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.<br />
<br />
There are 6 possible colors for the center triangle.<br />
<br />
*There are <math>{6\choose3} = 20</math> possible choices for the three outer triangles, if all three have different colors.<br />
*There are <math>6\cdot 5 = 30</math> (or <math>2 {6\choose2}</math>) possible choices for the three outer triangles, if two are one color and the third is a different color.<br />
*There are <math>{6\choose1} = 6</math> possible choices for the three outer triangles, if all three are the same color.<br />
<br />
Thus, in total we have <math>6\cdot(20 + 30 + 6) = 336</math> total possibilities.<br />
<br />
<br />
== Solution 2 (Easy burnsides) == <br />
We apply burnsides lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by <math>3+3=6</math> for our final count. <br />
<br />
Case 1: 0 degree rotation. This is known as the identity rotation, and there are <math>6^4=1296</math> choices because we don't have any restrictions.<br />
<br />
Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have <math>6*6=36</math> choices from this. <br />
<br />
Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have <math>6*6=36</math> choices from this.<br />
<br />
Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have <math>6*6*6=216</math> ways for one symmetry. There are 3 symmetries, so there are <math>216*3=648</math> combinations in all.<br />
<br />
<br />
Now, we add our cases up: <math>1296+36+36+648=2016</math>. We have to divide by 6, so <math>2016/6=\boxed{336}</math> distinguishable ways to color the triangle.<br />
<br />
== See also ==<br />
{{AIME box|year=2006|n=II|num-b=7|num-a=9}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_20&diff=896442012 AMC 10A Problems/Problem 202018-01-11T01:05:03Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #15]] and [[2012 AMC 10A Problems|2012 AMC 10A #20]]}}<br />
<br />
==Problem==<br />
<br />
A <math>3 \times 3</math> square is partitioned into <math>9</math> unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated <math>90\,^{\circ}</math> clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?<br />
<br />
<math> \textbf{(A)}\ \frac{49}{512}\qquad\textbf{(B)}\ \frac{7}{64}\qquad\textbf{(C)}\ \frac{121}{1024}\qquad\textbf{(D)}\ \frac{81}{512}\qquad\textbf{(E)}\ \frac{9}{32} </math><br />
<br />
== Solution ==<br />
First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is <math>1</math> case with all black squares. There are four cases with one white square and all <math>4</math> work. There are six cases with two white squares, but only the <math>2</math> with the white squares diagonal from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is <math>1+4+2=7</math>. In essence, the edges work the same way, so there are also <math>7</math> ways to color them. The number of ways to fit the conditions over the number of ways to color the squares is<br />
<br />
<cmath>\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}</cmath><br />
<br />
<br />
Solution 2<br />
We proceed by casework.<br />
Note that the middle square must be black, because when rotated 90 degrees, it must keep its position. Now we have to deal with the following cases: <br />
Case 1: 0 white squares.<br />
There is exactly <math> 1 </math> way to color the grid this way.<br />
Case 2: 1 white square.<br />
Note that the white square can be anywhere on the grid except for the middle square, because when rotating 90 degrees it can never land on it self. Thus, there are <math>8</math> cases.<br />
Case 3: 2 white squares. <br />
We have <math>\binom{8}{2}=28</math> ways to color two white squares without restrictions (the middle square must be black, giving us 8 squares to choose from). However, we must subtract the ways in which two white squares differ by a rotation of 90 degrees about the middle of the square. There are a total of 8 cases we must subtract (these are not too hard to see). Thus, there are <math>20</math> ways from this.<br />
Case 4: 3 white squares.<br />
Since we can not change the middle square, there are <math>binom{8}{3}=56</math> ways to color this. However, we must consider the cases where at least two squares differ by a rotation of 90 degrees. We can count this with PIE: by the Principle of Exclusion, the number of cases we want to exclude is the number of cases where 2 squares differ by a rotation of 90 degrees and minus when there are 3 squares such that two of them differ by rotation of 90 and 1 of them differs by rotation of 180, because of the overcount from our first case. <br />
From case 2, there are 8 causes such that two squares differ by a rotation of 90. There are also 6 other places we can place the third square (it can't be the middle of the two that we already colored), for a total of 48 ways. We have to subtract the second case. Note that there are 8 ways in which we can arrange two squares differing by 180 degrees. Out of these, each one has two ways to put another square such that two differ by 90 degrees and 1 pair differs by 180. However, this is overcounted by a factor of 2, so there are actually 8*2/2=8 ways. <br />
Thus, we have <math>56-(48-8)=16</math> ways in this case.<br />
Case 5: 4 white squares.<br />
Note that two of them have to be on one of the 4 corner squares, and two of them have to be on one of the 4 edge squares. Each solution yields two combinations, for a total of 2*2=4.<br />
Adding up our cases yields <math>1+8+20+16+4=49</math> ways. There are 512 ways to color the square without restrictions. Thus, the answer is <math>\boxed{49}{512}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2012|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_22&diff=895982010 AMC 10A Problems/Problem 222018-01-08T17:41:48Z<p>Gamemaster402: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math><br />
<br />
==Solution 1==<br />
<br />
To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only 1 way to connect the chords such that a triangle is formed in the circle's interior. Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>.<br />
<br />
==Solution 2==<br />
There are a total of <math>\dbinom{8}{3}=56</math> total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be <math>\boxed{\textbf{(A) }28}</math>.<br />
<br />
proposed by mumpu2k16<br />
<br />
In general, the number of triangles that are in the interior will always be <math>\binom{n}{6}</math> (can you prove it?)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_15&diff=887281993 AIME Problems/Problem 152017-11-30T18:57:03Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{CH}</math> be an altitude of <math>\triangle ABC</math>. Let <math>R\,</math> and <math>S\,</math> be the points where the circles inscribed in the triangles <math>ACH\,</math> and <math>BCH^{}_{}</math> are tangent to <math>\overline{CH}</math>. If <math>AB = 1995\,</math>, <math>AC = 1994\,</math>, and <math>BC = 1993\,</math>, then <math>RS\,</math> can be expressed as <math>m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime integers. Find <math>m + n\,</math>.<br />
<br />
== Solution ==<br />
<br />
<asy><br />
unitsize(48);<br />
pair A,B,C,H;<br />
A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);<br />
label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE);<br />
draw(circle((2,1),1));<br />
pair [] x=intersectionpoints(C--H,circle((2,1),1));<br />
dot(x[0]); label("$S$",x[0],SW);<br />
draw(circle((4.29843788128,1.29843788128),1.29843788128));<br />
pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128));<br />
dot(y[0]); label("$R$",y[0],NE);<br />
label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S);<br />
</asy><br />
<br />
From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2</math>. Subtracting those two equations yields <math>AH^2-(1995-AH)^2=3987</math>. After simplification, we see that <math>2*1995AH-1995^2=3987</math>, or <math>AH=\frac{1995}{2}+\frac{3987}{2*1995}</math>. Note that <math>AH+BH=1995</math>. Therefore we have that <math>BH=\frac{1995}{2}-\frac{3987}{2*1995}</math>. Therefore <math>AH-BH=\frac{3987}{1995}</math>.<br />
<br />
Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{AH+CH-AC}{2}</math>, and <math>HS=\frac{CH+BH-BC}{2}</math>. Therefore we have<br />
<br />
<math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</math>.<br />
<br />
Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>.<br />
<br />
Edit by GameMaster402: It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2}</math> Plugging in <math>n=1994</math> yields that the answer is <math>1992/1995*2</math>, which simplifies to <math>332/665</math><br />
<br />
== See also ==<br />
{{AIME box|year=1993|num-b=14|after=Last question}}<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_15&diff=887271993 AIME Problems/Problem 152017-11-30T18:53:59Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{CH}</math> be an altitude of <math>\triangle ABC</math>. Let <math>R\,</math> and <math>S\,</math> be the points where the circles inscribed in the triangles <math>ACH\,</math> and <math>BCH^{}_{}</math> are tangent to <math>\overline{CH}</math>. If <math>AB = 1995\,</math>, <math>AC = 1994\,</math>, and <math>BC = 1993\,</math>, then <math>RS\,</math> can be expressed as <math>m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime integers. Find <math>m + n\,</math>.<br />
<br />
== Solution ==<br />
<br />
<asy><br />
unitsize(48);<br />
pair A,B,C,H;<br />
A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);<br />
label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE);<br />
draw(circle((2,1),1));<br />
pair [] x=intersectionpoints(C--H,circle((2,1),1));<br />
dot(x[0]); label("$S$",x[0],SW);<br />
draw(circle((4.29843788128,1.29843788128),1.29843788128));<br />
pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128));<br />
dot(y[0]); label("$R$",y[0],NE);<br />
label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S);<br />
</asy><br />
<br />
From the [[Pythagorean Theorem]], <math>AH^2+CH^2=1994^2</math>, and <math>(1995-AH)^2+CH^2=1993^2</math>. Subtracting those two equations yields <math>AH^2-(1995-AH)^2=3987</math>. After simplification, we see that <math>2*1995AH-1995^2=3987</math>, or <math>AH=\frac{1995}{2}+\frac{3987}{2*1995}</math>. Note that <math>AH+BH=1995</math>. Therefore we have that <math>BH=\frac{1995}{2}-\frac{3987}{2*1995}</math>. Therefore <math>AH-BH=\frac{3987}{1995}</math>.<br />
<br />
Now note that <math>RS=|HR-HS|</math>, <math>RH=\frac{AH+CH-AC}{2}</math>, and <math>HS=\frac{CH+BH-BC}{2}</math>. Therefore we have<br />
<br />
<math>RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}</math>.<br />
<br />
Plugging in <math>AH-BH</math> and simplifying, we have <math>RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}</math>.<br />
<br />
Edit by GameMaster402: It can be shown that in any triangle with side lengths <math>n-1, n, n+1</math>, if you draw an altitude from the vertex to the side of <math>n+1</math>, and draw the incircles of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-3}{2n}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1993|num-b=14|after=Last question}}<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_23&diff=864452012 AMC 10A Problems/Problem 232017-07-19T07:49:01Z<p>Gamemaster402: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #19]] and [[2012 AMC 10A Problems|2012 AMC 10A #23]]}}<br />
<br />
== Problem ==<br />
<br />
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?<br />
<br />
<br />
<math> \text{(A)}\ 60\qquad\text{(B)}\ 170\qquad\text{(C)}\ 290\qquad\text{(D)}\ 320\qquad\text{(E)}\ 660 </math><br />
<br />
== Solution ==<br />
<br />
Note that if <math>n</math> is the number of friends each person has, then <math>n</math> can be any integer from <math>1</math> to <math>4</math>, inclusive.<br />
<br />
Also note that one person can only have at most 5 friends. That's right, 5 friends. xdxdxdxd 111111<br />
<br />
Also note that the cases of <math>n=1</math> and <math>n=4</math> are the same, since a map showing a solution for <math>n=1</math> can correspond one-to-one with a map of a solution for <math>n=4</math> by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with <math>n=2</math> when compared to configurations of <math>n=3</math>. Thus, we have two cases to examine, <math>n=1</math> and <math>n=2</math>, and we count each of these combinations twice.<br />
<br />
For <math>n=1</math>, if everyone has exactly one friend, that means there must be <math>3</math> pairs of friends, with no other interconnections. The first person has <math>5</math> choices for a friend. There are <math>4</math> people left. The next person has <math>3</math> choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are <math>15</math> configurations with <math>n=1</math>.<br />
<br />
For <math>n=2</math>, there are two possibilities. The group of <math>6</math> can be split into two groups of <math>3</math>, with each group creating a friendship triangle. The first person has <math>\binom{5}{2} = 10</math> ways to pick two friends from the other five, while the other three are forced together. Thus, there are <math>10</math> triangular configurations. <br />
<br />
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon [[without loss of generality]]. This person has <math>\binom{5}{2} = 10</math> choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are <math>10 \cdot 3! = 60</math> hexagonal configurations, and in total <math>70</math> configurations for <math>n=2</math>.<br />
<br />
As stated before, <math>n=3</math> has <math>70</math> configurations, and <math>n=4</math> has <math>15</math> configurations. This gives a total of <math>(70 + 15)\cdot 2 = 170</math> configurations, which is option <math>\boxed{\textbf{(B)}\ 170}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=22|num-a=24}}<br />
<br />
{{AMC12 box|year=2012|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Gamemaster402https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=855992016 AMC 10A Problems2017-05-09T14:46:09Z<p>Gamemaster402: </p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math><br />
<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
<br />
[[2016 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
<br />
[[2016 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
<br />
[[2016 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
<br />
[[2016 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
<br />
[[2016 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
<br />
[[2016 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
<br />
[[2016 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
<br />
[[2016 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
<br />
[[2016 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
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[[2016 AMC 10A Problems/Problem 15|Solution]]<br />
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==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
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<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
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<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
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<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
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<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
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<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
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[[2016 AMC 10A Problems/Problem 16|Solution]]<br />
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==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
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<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
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[[2016 AMC 10A Problems/Problem 17|Solution]]<br />
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==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
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<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
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[[2016 AMC 10A Problems/Problem 18|Solution]]<br />
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==Problem 19==<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
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<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
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<br />
[[2016 AMC 10A Problems/Problem 19|Solution]]<br />
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==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
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<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
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[[2016 AMC 10A Problems/Problem 20|Solution]]<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
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<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
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[[2016 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
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<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
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[[2016 AMC 10A Problems/Problem 23|Solution]]<br />
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==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
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<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
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[[2016 AMC 10A Problems/Problem 24|Solution]]<br />
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==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
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<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Gamemaster402